MCQ 511 Mark
$\int_{}^{} {\frac{{\cos x - \sin x}}{{1 + \sin 2x}}\;dx = } $
- ✓
$ - \frac{1}{{\cos x + \sin x}} + c$
- B
$\frac{1}{{\cos x + \sin x}} + c$
- C
$\frac{1}{{\cos x - \sin x}} + c$
- D
AnswerCorrect option: A. $ - \frac{1}{{\cos x + \sin x}} + c$
a
(a)$\int_{}^{} {\frac{{\cos x - \sin x}}{{1 + \sin 2x}}\,dx} = \int_{}^{} {\frac{{\cos x - \sin x}}{{{{(\sin x + \cos x)}^2}}}\,dx} $
Now put $\sin x + \cos x = t,$ then the $ \Rightarrow (\cos x - \sin x)\,dx = dt$
required integral is $ - \frac{1}{{\sin x + \cos x}} + c$.
View full question & answer→MCQ 521 Mark
$\int_{}^{} {\frac{1}{{x{{\cos }^2}(1 + \log x)}}\;dx = } $
- ✓
$\tan \,(1 + \log x) + c$
- B
$\cot \,(1 + \log x) + c$
- C
$ - \tan \,(1 + \log x) + c$
- D
$ - \cot (\,1 + \log x) + c$
AnswerCorrect option: A. $\tan \,(1 + \log x) + c$
a
(a) Put $1 + \log x = t \Rightarrow \frac{1}{x}\,dx = dt,$ then
$\int_{}^{} {\frac{1}{{x{{\cos }^2}(1 + \log x)}}\,dx} = \int_{}^{} {\frac{{dt}}{{{{\cos }^2}t}} = \int_{}^{} {{{\sec }^2}t\,dt} } $
$ = \tan t + c = \tan (1 + \log x) + c.$
View full question & answer→MCQ 531 Mark
$\int_{}^{} {\frac{1}{{{x^2}\sqrt {1 + {x^2}} }}} \;dx = $
- ✓
$ - \frac{{\sqrt {1 + {x^2}} }}{x} + c$
- B
$\frac{{\sqrt {1 + {x^2}} }}{x} + c$
- C
$ - \frac{{\sqrt {1 - {x^2}} }}{x} + c$
- D
$ - \frac{{\sqrt {{x^2} - 1} }}{x} + c$
AnswerCorrect option: A. $ - \frac{{\sqrt {1 + {x^2}} }}{x} + c$
a
(a) Put $x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta ,$ then
$\int_{}^{} {\frac{1}{{{x^2}\sqrt {1 + {x^2}} }}\,dx = \int_{}^{} {\frac{{{{\sec }^2}\theta \,d\theta }}{{{{\tan }^2}\theta \sec \theta }} = \int_{}^{} {{\rm{cosec}}\,\theta \,{\rm{cot}}\theta \,d\theta } } } $
$ = - {\rm{cosec}}\,\theta + {\rm{c}} = \frac{{{\rm{-}}\sqrt {{x^{\rm{2}}} + 1} }}{x} + c.$
View full question & answer→MCQ 541 Mark
$\int_{}^{} {\frac{{\log (x + \sqrt {1 + {x^2}} )}}{{\sqrt {1 + {x^2}} }}\;dx = } $
- ✓
$\frac{1}{2}{[\log (x + \sqrt {1 + {x^2}} )]^2} + c$
- B
$\log {(x + \sqrt {1 + {x^2}} )^2} + c$
- C
$\log (x + \sqrt {1 + {x^2}} ) + c$
- D
AnswerCorrect option: A. $\frac{1}{2}{[\log (x + \sqrt {1 + {x^2}} )]^2} + c$
a
(a) Put $\log (x + \sqrt {1 + {x^2}} ) = t \Rightarrow \frac{1}{{\sqrt {1 + {x^2}} }}\,dx = dt,$ then
$\int_{}^{} {\frac{{\log (x + \sqrt {1 + {x^2}} )}}{{\sqrt {1 + {x^2}} }}\,dx} = \int_{}^{} {t\,dt} $
$\int_{}^{} {\frac{{{t^2}}}{2}dt} $ $ = \frac{1}{2}{\left[ {\log (x + \sqrt {1 + {x^2}} )} \right]^2} + c$.
View full question & answer→MCQ 551 Mark
$\int_{}^{} {\frac{1}{{1 + {{\sin }^2}x}}\;dx = } $
- ✓
$\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\sqrt 2 \tan x) + k$
- B
$\sqrt 2 {\tan ^{ - 1}}(\sqrt 2 \tan x) + k$
- C
$ - \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\sqrt 2 \tan x) + k$
- D
$ - \sqrt 2 {\tan ^{ - 1}}(\sqrt 2 \tan x) + k$
AnswerCorrect option: A. $\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\sqrt 2 \tan x) + k$
a
(a) $I = \int_{}^{} {\frac{1}{{1 + {{\sin }^2}x}}\,dx} = \int_{}^{} {\frac{{dx}}{{2{{\sin }^2}x + {{\cos }^2}x}}} $
$ = \int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{2{{\tan }^2}x + 1}}} $$ = \frac{1}{2}\int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{{{\tan }^2}x + \frac{1}{2}}}} $
Put $\tan x = t \Rightarrow {\sec ^2}x\,dx = dt,$ then
$I = \frac{1}{2}\int_{}^{} {\frac{{dt}}{{{t^2} + \frac{1}{2}}} = \frac{1}{2}} .\frac{1}{{1\sqrt 2 }}{\tan ^{ - 1}}\frac{t}{{1\sqrt 2 }}$
$ = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\sqrt 2 \tan x) + k$.
View full question & answer→MCQ 561 Mark
$\int_{}^{} {\frac{{{x^2}{{\tan }^{ - 1}}{x^3}}}{{1 + {x^6}}}\;dx} $ is equal to
- A
${\tan ^{ - 1}}({x^3}) + c$
- ✓
$\frac{1}{6}{({\tan ^{ - 1}}{x^3})^2} + c$
- C
$ - \frac{1}{2}{({\tan ^{ - 1}}{x^3})^2} + c$
- D
$\frac{1}{2}{({\tan ^{ - 1}}{x^2})^3} + c$
AnswerCorrect option: B. $\frac{1}{6}{({\tan ^{ - 1}}{x^3})^2} + c$
b
(b) Put ${x^3} = t \Rightarrow dt = 3{x^2}\,dx$
$\int_{}^{} {\frac{{{x^2}{{\tan }^{ - 1}}{x^3}\,dx}}{{1 + {x^6}}}} = \frac{1}{3}\int_{}^{} {\frac{{{{\tan }^{ - 1}}t}}{{1 + {t^2}}}} \,dt$
Put $z = {\tan ^{ - 1}}t \Rightarrow dz = \frac{{dt}}{{1 + {t^2}}}$
$ = \frac{1}{3}\int_{}^{} {z\,dz} = \frac{1}{3}\frac{{{z^2}}}{2} = \frac{{{z^2}}}{6} = \frac{1}{6}{({\tan ^{ - 1}}{x^3})^2} + c$.
View full question & answer→MCQ 571 Mark
$\int_{}^{} {\frac{{{e^{2x}} + 1}}{{{e^{2x}} - 1}}\;dx} $ equals
- ✓
$\log ({e^x} - {e^{ - x}}) + c$
- B
$\log ({e^x} + {e^{ - x}}) + c$
- C
$\log ({e^{ - x}} - {e^x}) + c$
- D
$\log (1 - {e^{ - x}}) + c$
AnswerCorrect option: A. $\log ({e^x} - {e^{ - x}}) + c$
a
(a) $I = \int_{}^{} {\frac{{{e^{2x}} + 1}}{{{e^{2x}} - 1}}} = \int_{}^{} {\frac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}}} \,dx$
Put ${e^x} - {e^{ - x}} = t \Rightarrow ({e^x} + {e^{ - x}})\,dx = dt$
$\therefore \,\,\,I = \int_{}^{} {\frac{{dt}}{t}\,dt} = \log t + c = \log ({e^x} - {e^{ - x}}) + c$.
View full question & answer→MCQ 581 Mark
$\int_{}^{} {(x + 3){{({x^2} + 6x + 10)}^9}\;dx} $ equals
- ✓
$\frac{1}{{20}}{({x^2} + 6x + 10)^{10}} + c$
- B
$\frac{1}{{20}}{(x + 3)^2}{({x^2} + 6x + 10)^{10}} + c$
- C
$\frac{1}{{16}}{({x^2} + 6x + 10)^8} + c$
- D
$\frac{1}{{38}}{(x + 3)^{19}} + \frac{1}{2}(x + 3) + c$
AnswerCorrect option: A. $\frac{1}{{20}}{({x^2} + 6x + 10)^{10}} + c$
a
(a)$\int_{}^{} {(x + 3){{({x^2} + 6x + 10)}^9}dx} $
$ = \frac{1}{2}\int_{}^{} {(2x + 6){{({x^2} + 6x + 10)}^2}dx} $
$ = \frac{1}{2}\frac{{{{({x^2} + 6x + 10)}^{10}}}}{{10}} + c$$ = \frac{1}{{20}}{({x^2} + 6x + 10)^{10}} + c$.
View full question & answer→MCQ 591 Mark
$\int {\frac{{{{\sin }^3}2x}}{{{{\cos }^5}2x}}dx = } $
AnswerCorrect option: D. $\frac{1}{8}{\tan ^4}2x + C$
d
(d) $I = \int {\frac{{{{\sin }^3}2x}}{{{{\cos }^5}2x}}dx} $
==> $I = \int {\frac{{{{\sin }^3}2x}}{{{{\cos }^3}2x}}.\frac{1}{{{{\cos }^2}2x}}dx = \int {{{\tan }^3}2x.{{\sec }^2}2x\,dx.} } $
Putting tan $2x = t$ and $2{\sec ^2}2x\,dx = dt$, we get
$I = \int {{t^3}\frac{{dt}}{2} = \frac{1}{2}.\frac{{{t^4}}}{4} + C = \frac{1}{8}({{\tan }^4}2x) + C.} $
View full question & answer→MCQ 601 Mark
$\int {{e^{3\log x}}{{({x^4} + 1)}^{ - 1}}\,\,dx} $=
AnswerCorrect option: B. $\frac{1}{4}\log ({x^4} + 1) + c$
b
(b) $I = \int {{e^{3\log x}}{{({x^4} + 1)}^{ - 1}}dx} $$ = \int {{e^{\log {x^3}}}{{({x^4} + 1)}^{ - 1}}dx} $
$ = \frac{1}{4}\int {\frac{{4{x^3}}}{{({x^4} + 1)}}dx} = \frac{1}{4}\log ({x^4} + 1) + c$.
View full question & answer→MCQ 611 Mark
$\int_{}^{} {\frac{{dx}}{{2\sqrt x (1 + x)}} = } $
- A
$\frac{1}{2}{\tan ^{ - 1}}(\sqrt x ) + c$
- ✓
${\tan ^{ - 1}}(\sqrt x ) + c$
- C
$2{\tan ^{ - 1}}(\sqrt x ) + c$
- D
AnswerCorrect option: B. ${\tan ^{ - 1}}(\sqrt x ) + c$
b
(b) $I = \int {\frac{{dx}}{{2\sqrt x (1 + x)}}} $
Put $\sqrt x \, = t$==> $\frac{1}{{2\sqrt x }}dx = dt$
$\therefore I = \int {\frac{{dt}}{{1 + {t^2}}}} = {\tan ^{ - 1}}t + c$$ = {\tan ^{ - 1}}(\sqrt x ) + c$.
View full question & answer→MCQ 621 Mark
$\int_{}^{} {\frac{{\sqrt x }}{{1 + x}}dx = } $
- A
$\sqrt x - {\tan ^{ - 1}}\sqrt x + c$
- ✓
$2(\sqrt x - {\tan ^{ - 1}}\sqrt x ) + c$
- C
$2(\sqrt x + {\tan ^{ - 1}}x) + c$
- D
$\sqrt {1 + x} + c$
AnswerCorrect option: B. $2(\sqrt x - {\tan ^{ - 1}}\sqrt x ) + c$
b
(b)$\int_{}^{} {\frac{{\sqrt x }}{{1 + x}}\,dx = \int_{}^{} {\frac{{\sqrt x \sqrt x }}{{\sqrt x (1 + x)}}\,dx} } $
$ = \int_{}^{} {\frac{{x + 1}}{{\sqrt x (x + 1)}}\,dx - \int_{}^{} {\frac{1}{{\sqrt x (x + 1)}}\,dx} } $
$ = \int_{}^{} {\frac{1}{{\sqrt x }}\,dx - \int_{}^{} {\frac{1}{{\sqrt x (x + 1)}}\,dx} } $
$ = 2{x^{1/2}} - 2{\tan ^{ - 1}}\sqrt x + c = 2(\sqrt x - {\tan ^{ - 1}}\sqrt x ) + c.$
View full question & answer→MCQ 631 Mark
$\int_{}^{} {\frac{{\sin x}}{{\sin x - \cos x}}} \;dx = $
- A
$\frac{1}{2}\log (\sin x - \cos x) + x + c$
- ✓
$\frac{1}{2}[\log (\sin x - \cos x) + x] + c$
- C
$\frac{1}{2}\log (\cos x - \sin x) + x + c$
- D
$\frac{1}{2}[\log (\cos x - \sin x) + x] + c$
AnswerCorrect option: B. $\frac{1}{2}[\log (\sin x - \cos x) + x] + c$
b
(b)$\int_{}^{} {\frac{{\sin x\,dx}}{{\sin x - \cos x}}} = \frac{1}{2}\int_{}^{} {\frac{{2\sin x}}{{\sin x - \cos x}}\,dx} $
$ = \frac{1}{2}\int_{}^{} {\frac{{(\sin x - \cos x + \sin x + \cos x)}}{{\sin x - \cos x}}\,dx} $
$ = \frac{1}{2}\int_{}^{} {\left( {1 + \frac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right)\,dx} $
$= \frac{1}{2}[x + \log (\sin x - \cos x)] + c$.
View full question & answer→MCQ 641 Mark
The value of $\int_{}^{} {\frac{{dx}}{{\sqrt x \,(x + 9)}}dx} $ is equal to
- A
${\tan ^{ - 1}}\sqrt x $
- B
${\tan ^{ - 1}}\left( {\frac{{\sqrt x }}{3}} \right)$
- C
$\frac{2}{3}{\tan ^{ - 1}}\sqrt x $
- ✓
$\frac{2}{3}{\tan ^{ - 1}}\left( {\frac{{\sqrt x }}{3}} \right)$
AnswerCorrect option: D. $\frac{2}{3}{\tan ^{ - 1}}\left( {\frac{{\sqrt x }}{3}} \right)$
d
(d) We have, $I = \int_{}^{} {\frac{{dx}}{{\sqrt x (x + 9)}}} $
Put $\sqrt x = t$, squaring both sides, we get $x = {t^2}$ and $dx = 2tdt$
$\therefore $$I = 2\int_{}^{} {\frac{{dt}}{{{t^2} + {3^2}}}} = \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{t}{3}} \right)$ ==> $I = \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{{\sqrt x }}{3}} \right)$.
View full question & answer→MCQ 651 Mark
$\int_{}^{} {\frac{{\sin 2xdx}}{{1 + {{\cos }^2}x}}} = $
- A
$\frac{1}{2}\log (1 + {\cos ^2}x) + c$
- B
$2\log (1 + {\cos ^2}x) + c$
- C
$\frac{1}{2}\log (1 + \cos 2x) + c$
- ✓
$ - \log (1 + {\cos ^2}x) + c$
AnswerCorrect option: D. $ - \log (1 + {\cos ^2}x) + c$
d
(d) $I = \int {\frac{{\sin 2x}}{{1 + {{\cos }^2}x}}dx = \int {\frac{{2\sin x\cos x}}{{1 + {{\cos }^2}x}}dx} } $
Put $1 + {\cos ^2}x = t$ ==> $ - 2\sin x\cos x\,\,dx = dt$
==> $\sin 2x = - dt$. Hence
$I = \int {^ - \left( {\frac{{dt}}{t}} \right)} = - \log t + c$
$ = - \log (1 + {\cos ^2}x) + c$.
View full question & answer→MCQ 661 Mark
${I_1} = \int {{{\sin }^{ - 1}}x\,\,dx} $ and ${I_2} = \int {{{\sin }^{ - 1}}\sqrt {1 - {x^2}} } dx$then
AnswerCorrect option: C. ${I_1} + {I_2} = \frac{\pi }{2}x$
c
(c) ${I_1} = \int {{{\sin }^{ - 1}}xdx} $
Let ${\sin ^{ - 1}}x = \theta $==> $x = \sin \theta $ ==> $dx = \cos \theta \,d\theta $
${I_1} = \int {\theta \cos \theta d\theta } $$ = \theta \sin \theta - \int {\sin \theta d\theta } $$ = \theta \sin \theta + \cos \theta $
$ = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} $
${I_2} = \int {{{\sin }^{ - 1}}\sqrt {1 - {x^2}} } dx$$ = \int {{{\cos }^{ - 1}}xdx} $
Let $\cos \phi = x,$ Hence $ - \sin \phi \,d\phi = dx$
${I_2} = - \int {\phi hi \sin \phi d\phi } $$ = \phi \cos \phi + \int { - \cos \phi d\phi } $
$ = \phi \cos \phi - \sin \phi $$ = x{\cos ^{ - 1}}x - \sqrt {1 - {x^2}} $
${I_1} + {I_2} = x({\cos ^{ - 1}}x + {\sin ^{ - 1}}x) = \frac{\pi }{2}x$.
View full question & answer→MCQ 671 Mark
$\int_{}^{} {{{\sin }^3}{\kern 1pt} x{{\cos }^2}x\;dx = } $
- ✓
$\frac{{{{\cos }^5}x}}{5} - \frac{{{{\cos }^3}x}}{3} + c$
- B
$\frac{{{{\cos }^5}x}}{5} + \frac{{{{\cos }^3}x}}{3} + c$
- C
$\frac{{{{\sin }^5}x}}{5} - \frac{{{{\sin }^3}x}}{3} + c$
- D
$\frac{{{{\sin }^5}x}}{5} + \frac{{{{\sin }^3}x}}{3} + c$
AnswerCorrect option: A. $\frac{{{{\cos }^5}x}}{5} - \frac{{{{\cos }^3}x}}{3} + c$
a
(a)$\int_{}^{} {{{\sin }^3}x{{\cos }^2}x\,dx} = \int_{}^{} {(1 - {{\cos }^2}x){{\cos }^2}x\,.\,\sin x\,dx} $
Put $\cos x = t \Rightarrow - \sin x\,dx = dt,$ then it reduces to
$ - \int_{}^{} {({t^2} - {t^4})dt} = \frac{{{t^5}}}{5} - \frac{{{t^3}}}{3} + c = \frac{{{{(\cos x)}^5}}}{5} - \frac{{{{(\cos x)}^3}}}{3} + c$.
View full question & answer→MCQ 681 Mark
$\int_{}^{} {{{\sin }^5}x{{\cos }^4}x\;dx = } $
- ✓
$ - \frac{1}{5}{\cos ^5}x + \frac{2}{7}{\cos ^7}x - \frac{1}{9}{\cos ^9}x + c$
- B
$\frac{1}{5}{\cos ^5}x + \frac{2}{7}{\cos ^7}x - \frac{1}{9}{\cos ^9}x + c$
- C
$\frac{1}{5}{\cos ^5}x + \frac{2}{7}{\cos ^7}x + \frac{1}{9}{\cos ^9}x + c$
- D
AnswerCorrect option: A. $ - \frac{1}{5}{\cos ^5}x + \frac{2}{7}{\cos ^7}x - \frac{1}{9}{\cos ^9}x + c$
a
(a) Put $\cos x = t \Rightarrow - \sin x\,dx = dt,$ then
$\int_{}^{} {{{(1 - {{\cos }^2}x)}^2}.{{\cos }^4}x\sin x\,dx} = - \int_{}^{} {{{(1 - {t^2})}^2}.\,{t^4}dt} $
$ = - \frac{{{t^5}}}{5} + \frac{2}{7}{t^7} - \frac{1}{9}{t^9} + c = - \frac{{{{\cos }^5}x}}{5} + \frac{2}{7}{\cos ^7}x - \frac{1}{9}{\cos ^9}x + c$.
Aliter : By reduction formula.
View full question & answer→MCQ 691 Mark
$\int_{}^{} {\frac{{dx}}{{x({x^7} + 1)}}} = $
- A
$\log \left( {\frac{{{x^7}}}{{{x^7} + 1}}} \right) + c$
- ✓
$\frac{1}{7}\log \left( {\frac{{{x^7}}}{{{x^7} + 1}}} \right) + c$
- C
$\log \left( {\frac{{{x^7} + 1}}{{{x^7}}}} \right) + c$
- D
$\frac{1}{7}\log \left( {\frac{{{x^7} + 1}}{{{x^7}}}} \right) + c$
AnswerCorrect option: B. $\frac{1}{7}\log \left( {\frac{{{x^7}}}{{{x^7} + 1}}} \right) + c$
b
(b) Given, $\int_{}^{} {\frac{{dx}}{{x\,({x^7} + 1)}}} = \int_{}^{} {\frac{{dx}}{{{x^8}\left( {1 + \frac{1}{{{x^7}}}} \right)}}} $
Put $1 + \frac{1}{{{x^7}}} = t$ ==> $\frac{{ - 7}}{{{x^8}}}dx = dt$
$I = \frac{{ - 1}}{7}\int {\frac{{dt}}{t} = } \frac{{ - 1}}{7}\log t + c$
==> $I = - \frac{1}{7}\log \left( {\frac{{{x^7} + 1}}{{{x^7}}}} \right) + c$ ==> $I = \frac{1}{7}\log \left( {\frac{{{x^7}}}{{{x^7} + 1}}} \right) + c$.
View full question & answer→MCQ 701 Mark
$\int_{}^{} {\frac{{dx}}{{x({x^5} + 1)}}} = $
- A
$\frac{1}{5}\log {x^5}({x^5} + 1) + c$
- B
$\frac{1}{5}\log {x^5}\left( {\frac{{1 + {x^5}}}{{{x^5}}}} \right) + c$
- C
$\frac{1}{5}\log {x^5}\left( {\frac{{{x^5}}}{{{x^5} + 1}}} \right) + c$
- ✓
Answerd
(d) We have $I = \int {\frac{{dx}}{{x({x^5} + 1)}}} = \int {\frac{{dx}}{{{x^6}\left( {1 + \frac{1}{{{x^5}}}} \right)}}} $
Put $1 + \frac{1}{{{x^5}}} = t$ ==> $\frac{{ - 5}}{{{x^6}}}dx = dt$
==> $I = - \frac{1}{5}\int {\frac{{dt}}{t} = - \frac{1}{5}} \log t + c$
$I = - \frac{1}{5}\log \left( {1 + \frac{1}{{{x^5}}}} \right) + c = - \frac{1}{5}\log \left( {\frac{{{x^5} + 1}}{{{x^5}}}} \right) + c$
$I = \frac{1}{5}\log \left( {\frac{{{x^5}}}{{{x^5} + 1}}} \right) + c$.
View full question & answer→MCQ 711 Mark
$\int {\frac{{1 + x + \sqrt {x + {x^2}} }}{{\sqrt x \, + \sqrt {1 + x} }}\,\,dx = } $
- A
$1/2\sqrt {1 + x} + c$
- ✓
$2/3{(1 + x)^{3/2}} + c$
- C
$\sqrt {1 + x} + c$
- D
$2{(1 + x)^{3/2}} + c$
AnswerCorrect option: B. $2/3{(1 + x)^{3/2}} + c$
b
(b) $\int {\frac{{1 + x + \sqrt {x + {x^2}} }}{{\sqrt x + \sqrt {1 + x} }}dx} $$ = \int {\frac{{\sqrt {1 + x} [\sqrt {1 + x} + \sqrt x ]}}{{(\sqrt x + \sqrt {1 + x} )}}\,dx} $
$ = \int {\sqrt {1 + x\,} } dx = \frac{2}{3}{(1 + x)^{3/2}} + c$.
View full question & answer→MCQ 721 Mark
$\int_{}^{} {\frac{{dx}}{{(1 + {x^2})\sqrt {1 - {x^2}} }} = } $
- A
$\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 - {x^2}} }}{{x\sqrt 2 }}} \right] + c$
- ✓
$\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left[ {\frac{{x\sqrt 2 }}{{\sqrt {1 - {x^2}} }}} \right] + c$
- C
$\sqrt 2 {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 - {x^2}} }}{{x\sqrt 2 }}} \right] + c$
- D
$ - \sqrt 2 {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 - {x^2}} }}{{x\sqrt 2 }}} \right] + c$
AnswerCorrect option: B. $\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left[ {\frac{{x\sqrt 2 }}{{\sqrt {1 - {x^2}} }}} \right] + c$
b
(b) Put $x = \sin \theta \Rightarrow dx = \cos \theta \,d\theta ,$ then
$\int_{}^{} {\frac{{dx}}{{(1 + {x^2})\sqrt {1 - {x^2}} }} = \int_{}^{} {\frac{1}{{1 + {{\sin }^2}\theta }}\,d\theta } } = \int_{}^{} {\frac{{{{\sec }^2}\theta }}{{1 + 2{{\tan }^2}\theta }}\,d\theta } $
Again put $t = \tan \theta \Rightarrow dt = {\sec ^2}\theta \,d\theta ,$ then it reduces to
$\int_{}^{} {\frac{1}{{1 + 2{t^2}}}\,dt} = \frac{1}{2}\int_{}^{} {\frac{1}{{{t^2} + {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}\,dt} $
$ = \frac{1}{2}\left( {\frac{1}{{(1/\sqrt 2 )}}} \right){\tan ^{ - 1}}\left( {\frac{t}{{(1/\sqrt 2 )}}} \right) + c$
$ = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\sqrt 2 \tan \theta ) + c = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{x\sqrt 2 }}{{\sqrt {1 - {x^2}} }}} \right) + c.$
Aliter : Put first $x = \frac{1}{t}$ and then ${t^2} - 1 = {z^2}.$
View full question & answer→MCQ 731 Mark
$\int_{}^{} {\sqrt {\frac{{a + x}}{{a - x}}} \;dx = } $
- A
$a = \frac{1}{2}$
- B
$a{\cos ^{ - 1}}x/a - \sqrt {{a^2} - {x^2}} + c$
- C
$ - a{\cos ^{ - 1}}x/a + \sqrt {{a^2} - {x^2}} + c$
- ✓
$ - a{\cos ^{ - 1}}x/a - \sqrt {{a^2} - {x^2}} + c$
AnswerCorrect option: D. $ - a{\cos ^{ - 1}}x/a - \sqrt {{a^2} - {x^2}} + c$
d
(d) $\int_{}^{} {\sqrt {\frac{{a + x}}{{a - x}}} \,dx} $. Put $x = a\cos \theta $
$ \Rightarrow dx = - a\sin \theta \,d\theta ,$ then it reduces to
$ - a\int_{}^{} {\sqrt {\frac{{1 + \cos \theta }}{{1 - \cos \theta }}} } (\sin \theta )\,d\theta $
$ = - 2a\int_{}^{} {\sqrt {\frac{{2{{\cos }^2}(\theta /2)}}{{2{{\sin }^2}(\theta /2)}}} } \,.\,\sin \frac{\theta }{2}\cos \frac{\theta }{2}\,d\theta $
$ = - a\int_{}^{} {(1 + \cos \theta )\,d\theta } = - a\,\,\left[ {{{\cos }^{ - 1}}\frac{x}{a} + \sqrt {\frac{{{a^2} - {x^2}}}{a}} } \right] + c$
$ = - a{\cos ^{ - 1}}\frac{x}{a} - \sqrt {{a^2} - {x^2}} + c$.
View full question & answer→MCQ 741 Mark
$\int_{}^{} {\frac{{dx}}{{{{(2\sin x + \cos x)}^2}}}} = $
- A
$\frac{1}{2}\left( {\frac{1}{{2\tan x + 1}}} \right) + c$
- B
$\frac{1}{2}\log (2\tan x + 1) + c$
- ✓
$\frac{1}{{2 + \cot x}} + c$
- D
$ - \frac{1}{2}\left( {\frac{1}{{2\tan x - 1}}} \right) + c$
AnswerCorrect option: C. $\frac{1}{{2 + \cot x}} + c$
c
(c) $\int_{}^{} {\frac{{dx}}{{{{(2\sin x + \cos x)}^2}}} = \int_{}^{} {\frac{{dx}}{{{{\sin }^2}x{{(2 + \cot x)}^2}}}} } = \int_{}^{} {\frac{{{\rm{cose}}{{\rm{c}}^{\rm{2}}}x\,dx}}{{{{(2 + \cot x)}^2}}}} $
Put $(2 + \cot x) = t \Rightarrow - {\rm{cose}}{{\rm{c}}^{\rm{2}}}x\,dx = dt$
$ = \int_{}^{} {\frac{{ - dt}}{{{t^2}}}} = \frac{1}{t} + c = \frac{1}{{2 + \cot x}} + c.$
View full question & answer→MCQ 751 Mark
If $c$ is any arbitrary constant, then $\int {{2^{{2^{{2^x}}}}}{2^{{2^x}}}{2^x}dx} $ is equal to
- A
$\frac{{{2^{{2^x}}}}}{{{{(\ln 2)}^3}}} + c$
- ✓
$\frac{{{2^{{2^{{2^x}}}}}}}{{{{(\ln 2)}^3}}} + c$
- C
${2^{{2^{{2^x}}}}}{(\ln 2)^3} + c$
- D
AnswerCorrect option: B. $\frac{{{2^{{2^{{2^x}}}}}}}{{{{(\ln 2)}^3}}} + c$
b
(b) Putting ${2^{{2^{{2^x}}}}} = t \Rightarrow {2^{{2^{{2^x}}}}}{2^{{2^x}}}{2^x}{(\log 2)^3}dx = dt,$ we get
$\int_{}^{} {{2^{{2^{{2^x}}}}}{{.2}^{{2^x}}}{{.2}^x}dx} = \frac{1}{{{{(\log 2)}^3}}}\int_{}^{} {1\,.\,dt} = \frac{t}{{{{(\log 2)}^3}}} + c$
$ = \frac{{{2^{{2^{{2^x}}}}}}}{{{{(\log 2)}^3}}} + c$.
View full question & answer→MCQ 761 Mark
$\int {\frac{{dx}}{{1 - \cos x - \sin x}}} = $
- A
$\log |1 + \cot x/2| + c$
- B
$\log |1 - \tan x/2| + c$
- ✓
$\log |1 - \cot x/2| + c$
- D
$\log |1 + \tan x/2| + c$
AnswerCorrect option: C. $\log |1 - \cot x/2| + c$
c
(c)$I = \int {\frac{{dx}}{{1 - \cos x - \sin x}}} $
$ = \int {\frac{{dx}}{{1 - \frac{{[(1 - {{\tan }^2}(x/2)]}}{{[(1 + {{\tan }^2}(x/2)]}} - \frac{{2\tan (x/2)}}{{1 + {{\tan }^2}(x/2)}}}}} $
$ = \int {\frac{{{{\sec }^2}(x/2).dx}}{{1 + {{\tan }^2}(x/2) - 1 + {{\tan }^2}(x/2) - 2\tan (x/2)}}} $
$ = \int {\frac{{{{\sec }^2}(x/2)\,dx}}{{2{{\tan }^2}(x/2) - 2\tan (x/2)}}} $ $ = \int {\frac{{\frac{1}{2}.{{\sec }^2}\left( {\frac{x}{2}} \right)\,dx}}{{{{\tan }^2}\left( {\frac{x}{2}} \right) - \tan \left( {\frac{x}{2}} \right)}}} $
Put $\tan (x/2) = t$==> $\frac{1}{2}{\sec ^2}(x/2)\,dx = dt$
$I = \int {\frac{{dt}}{{{t^2} - t}}} $ $ = \int {\frac{{dt}}{{t\,(t - 1)}}} $$ = \int {\left[ {\frac{1}{{t - 1}} - \frac{1}{t}} \right]\,dt} $ ( Put tan $x = t,$ $\therefore {\sec ^2}x{\rm{ }}dx = dt$)
$ = \int {\frac{{dt}}{{t - 1}} - \int {\frac{{dt}}{t}} } = \log (t - 1) - \log t + c = \log \left| {\frac{{t - 1}}{t}} \right| + c$
$ = \log \left| {\,\frac{{\tan (x/2) - 1}}{{\tan (x/2)}}\,} \right| + c$$ = \log \left| {\,1 - \cot \,\frac{x}{2}\,} \right| + c$.
View full question & answer→MCQ 771 Mark
$\int {\frac{{{3^x}}}{{\sqrt {{9^x} - 1} }}\,\,dx = } $
- ✓
$\frac{1}{{\log 3}}\log |{3^x} + \sqrt {{9^x} - 1} | + c$
- B
$\frac{1}{{\log 3}}\log |{9^x} + \sqrt {{9^x} - 1} \,| + \,c$
- C
$\frac{1}{{\log 9}}\log |{3^x} + \sqrt {{9^x} - 1} \,| + \,c$
- D
$\frac{1}{{\log 9}}\log |\,{3^x} - \sqrt {{9^x} - 1} |\, + c$
AnswerCorrect option: A. $\frac{1}{{\log 3}}\log |{3^x} + \sqrt {{9^x} - 1} | + c$
a
(a)$I = \int {\frac{{{3^x}}}{{\sqrt {{9^x} - 1} }}\,dx} $$ = \int {\frac{{{3^x}dx}}{{\sqrt {\,{{({3^x})}^2} - \,1} }}} $ Put ${3^x} = t$==> ${3^x}\log 3\,dx = dt$ ==> ${3^x}\,dx = dt/\log 3$ $ \Rightarrow I = \frac{1}{{\log 3}}\int {\frac{{dt}}{{\sqrt {{t^2} - 1} }}} $$ = \frac{1}{{\log 3}}\log \left[ {t + \sqrt {{t^2} - 1} } \right] + c$ $ = \frac{1}{{{{\log }_e}3}}\log \left[ {{3^x} + \sqrt {{9^x} - 1} } \right] + c$.
View full question & answer→MCQ 781 Mark
$\int_{}^{} {\frac{{{e^x}dx}}{{\sqrt {a + b{e^x}} }}} = $
- ✓
$2/b\sqrt {a + b{e^x}} + c$
- B
$2b\sqrt {a + b{e^x}} + c$
- C
$\frac{1}{{2b}}\sqrt {a + b{e^x}} $
- D
$\frac{a}{b}\sqrt {a + b{e^x}} + c$
AnswerCorrect option: A. $2/b\sqrt {a + b{e^x}} + c$
a
(a) Put $a + b{e^x} = t \Rightarrow b{e^x}dx = dt,$ then
$\int_{}^{} {\frac{{{e^x}dx}}{{\sqrt {a + b{e^x}} }}\, = \frac{1}{b}\int_{}^{} {\frac{1}{{\sqrt t }}\,dt = \frac{2}{b}\sqrt t + c} } $$ = \frac{{2\sqrt {a + b{e^x}} }}{b} + c.$
View full question & answer→MCQ 791 Mark
$\int_{}^{} {\frac{{{x^2} - 1}}{{{x^4} + {x^2} + 1}}dx} $ is equal to
- A
$\log ({x^4} + {x^2} + 1) + c$
- ✓
$\frac{1}{2}\log \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}} + c$
- C
$\frac{1}{2}\log \frac{{{x^2} + x + 1}}{{{x^2} - x + 1}} + c$
- D
$\log \frac{{{x^2} - x + 1}}{{x + x + 1}} + c$
AnswerCorrect option: B. $\frac{1}{2}\log \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}} + c$
b
(b) $I = \int_{}^{} {\frac{{{x^2} - 1}}{{{x^4} + {x^2} + 1}}} \,dx$$ = \int_{}^{} {\frac{{{x^2}\left( {1 - \frac{1}{{{x^2}}}} \right)}}{{{x^2}\left[ {{{\left( {x + \frac{1}{x}} \right)}^2} - 1} \right]}}\,dx} $
Put $\left( {x + \frac{1}{x}} \right) = t \Rightarrow \left( {1 - \frac{1}{{{x^2}}}} \right)\,dx = dt$
$I = \int_{}^{} {\frac{{dt}}{{{t^2} - 1}}} = \frac{1}{2}\log \left| {\,\frac{{t - 1}}{{t + 1}}\,} \right| + c$
$\therefore $ $I = \frac{1}{2}\log \,\left| {\frac{{{x^2} - x + 1}}{{{x^2} + x + 1}}} \right| + c$ ==> $a = \frac{1}{2},\,b = \frac{1}{2}$.
View full question & answer→MCQ 801 Mark
$\int_{}^{} {\frac{{{x^2} + x - 6}}{{(x - 2)(x - 1)}}dx = } $
- A
$x + 2\log (x - 1) + c$
- B
$2x + 2\log (x - 1) + c$
- C
$x + 4\log (1 - x) + c$
- ✓
$x + 4\log (x - 1) + c$
AnswerCorrect option: D. $x + 4\log (x - 1) + c$
d
(d)$\int_{}^{} {\frac{{{x^2} + x - 6}}{{(x - 2)(x - 1)}}\,dx} = \int_{}^{} {\frac{{(x + 3)(x - 2)}}{{(x - 2)(x - 1)}}\,dx} = \int_{}^{} {\frac{{x + 3}}{{x - 1}}\,dx} $
$ = \int_{}^{} {\frac{{x - 1}}{{x - 1}}\,dx + \int_{}^{} {\frac{4}{{x - 1}}\,dx} } = x + 4\log (x - 1) + c$.
View full question & answer→MCQ 811 Mark
$\int_{}^{} {\frac{{x - 1}}{{(x - 3)(x - 2)}}dx = } $
- A
$\log (x - 3) - \log (x - 2) + c$
- ✓
$\log {(x - 3)^2} - \log (x - 2) + c$
- C
$\log (x - 3) + \log (x - 2) + c$
- D
$\log {(x - 3)^2} + \log (x - 2) + c$
AnswerCorrect option: B. $\log {(x - 3)^2} - \log (x - 2) + c$
b
(b)$\int_{}^{} {\frac{{x - 1}}{{(x - 3)(x - 2)}}\,dx} $
$ = \int_{}^{} {\frac{{x - 3}}{{(x - 3)(x - 2)}}\,dx + \int_{}^{} {\frac{2}{{(x - 3)(x - 2)}}} } \,dx$
$ = \log \left[ {\frac{{(x - 2){{(x - 3)}^2}}}{{{{(x - 2)}^2}}}} \right] + c = \log \left[ {\frac{{{{(x - 3)}^2}}}{{(x - 2)}}} \right] + c.$
Trick : By inspection, $\frac{d}{{dx}}\left\{ {\log (x - 3) - \log (x - 2)} \right\}$
$ = \frac{1}{{x - 3}} - \frac{1}{{x - 2}} = \frac{1}{{(x - 3)(x - 2)}}$
$ \Rightarrow \frac{d}{{dx}}\left\{ {2\log (x - 3) - \log (x - 2)} \right\}$
$ = \frac{2}{{x - 3}} - \frac{1}{{x - 2}} = \frac{{x - 1}}{{(x - 3)(x - 2)}}$.
View full question & answer→MCQ 821 Mark
$\int_{}^{} {\frac{1}{{\cos x(1 + \cos x)}}} \;dx = $
- A
$\log (\sec x + \tan x) + 2\tan \frac{x}{2} + c$
- B
$\log (\sec x + \tan x) - 2\tan \frac{x}{2} + c$
- C
$\log (\sec x + \tan x) + \tan \frac{x}{2} + c$
- ✓
$\log (\sec x + \tan x) - \tan \frac{x}{2} + c$
AnswerCorrect option: D. $\log (\sec x + \tan x) - \tan \frac{x}{2} + c$
d
(d)$\int_{}^{} {\frac{1}{{\cos x(1 + \cos x)}}} dx = \int_{}^{} {\frac{{dx}}{{\cos x}} - \int_{}^{} {\frac{{dx}}{{1 + \cos x}}} } $
$ = \int_{}^{} {\sec x\;dx - \frac{1}{2}\int_{}^{} {{{\sec }^2}\frac{x}{2}dx} } $
$ = \log (\sec x + \tan x) - \tan \frac{x}{2} + c.$
View full question & answer→MCQ 831 Mark
$\int_{}^{} {\frac{1}{{(x - 1)({x^2} + 1)}}dx} = $
- ✓
$\frac{1}{2}\log (x - 1) - \frac{1}{4}\log ({x^2} + 1) - \frac{1}{2}{\tan ^{ - 1}}x + c$
- B
$\frac{1}{2}\log (x - 1) + \frac{1}{4}\log ({x^2} + 1) - \frac{1}{2}{\tan ^{ - 1}}x + c$
- C
$\frac{1}{2}\log (x - 1) - \frac{1}{2}\log ({x^2} + 1) - \frac{1}{2}{\tan ^{ - 1}}x + c$
- D
AnswerCorrect option: A. $\frac{1}{2}\log (x - 1) - \frac{1}{4}\log ({x^2} + 1) - \frac{1}{2}{\tan ^{ - 1}}x + c$
a
(a) We have $\frac{1}{{(x - 1)({x^2} + 1)}} = \frac{A}{{(x - 1)}} + \frac{{Bx + C}}{{({x^2} + 1)}}$
$ \Rightarrow 1 = A({x^2} + 1) + (Bx + C)(x - 1)$
If $x = 1,$ then $A = \frac{1}{2}$ .....$(i)$
$A - C = 1 \Rightarrow C = - \frac{1}{2}$ .....$(ii)$
$A + B = 0 \Rightarrow B = - \frac{1}{2}$ .....$(iii)$
Putting these values, we get
$\frac{1}{{(x - 1)({x^2} + 1)}} = \frac{1}{2}.\frac{1}{{(x - 1)}} - \frac{{x + 1}}{{2({x^2} + 1)}}$
Hence $\int_{}^{} {\frac{1}{{(x - 1)({x^2} + 1)}}} dx = \frac{1}{2}\int_{}^{} {\frac{{dx}}{{(x - 1)}} - \frac{1}{2}} \,\int_{}^{} {\frac{{x + 1}}{{{x^2} + 1}}} dx$
$ = \frac{1}{2}\log (x - 1) - \frac{1}{4}\log ({x^2} + 1) - \frac{1}{2}{\tan ^{ - 1}}x + c.$
View full question & answer→MCQ 841 Mark
$\int_{}^{} {\frac{{{x^2} + x - 1}}{{{x^2} + x - 6}}\;dx = } $
- A
$x + \log (x + 3) + \log (x - 2) + c$
- ✓
$x - \log (x + 3) + \log (x - 2) + c$
- C
$x - \log (x + 3) - \log (x - 2) + c$
- D
AnswerCorrect option: B. $x - \log (x + 3) + \log (x - 2) + c$
b
(b)$\int_{}^{} {\frac{{{x^2} + x - 1}}{{{x^2} + x - 6}}\,dx} = \int_{}^{} {\left[ {1 + \frac{5}{{{x^2} + x - 6}}} \right]} \,dx$
$ = \int_{}^{} {\left[ {1 + \frac{5}{{(x + 3)(x - 2)}}} \right]} \,dx$$ = \int_{}^{} {dx} + \int_{}^{} {\frac{{dx}}{{x - 2}}} - \int_{}^{} {\frac{{dx}}{{x + 3}}} $
$ = x + \log (x - 2) - \log (x + 3) + c$.
View full question & answer→MCQ 851 Mark
$\int_{}^{} {\frac{{{x^2}}}{{({x^2} + 2)({x^2} + 3)}}\;} dx = $
- A
$ - \sqrt 2 {\tan ^{ - 1}}x + \sqrt 3 {\tan ^{ - 1}}x + c$
- ✓
$ - \sqrt 2 {\tan ^{ - 1}}\frac{x}{{\sqrt 2 }} + \sqrt 3 {\tan ^{ - 1}}\frac{x}{{\sqrt 3 }} + c$
- C
$\sqrt 2 {\tan ^{ - 1}}\frac{x}{{\sqrt 2 }} + \sqrt 3 {\tan ^{ - 1}}\frac{x}{{\sqrt 3 }} + c$
- D
AnswerCorrect option: B. $ - \sqrt 2 {\tan ^{ - 1}}\frac{x}{{\sqrt 2 }} + \sqrt 3 {\tan ^{ - 1}}\frac{x}{{\sqrt 3 }} + c$
b
(b)$\int_{}^{} {\frac{{{x^2}}}{{({x^2} + 2)({x^2} + 3)}}} \,dx = \int_{}^{} {\left[ {\frac{3}{{{x^2} + 3}} - \frac{2}{{{x^2} + 2}}} \right]} \,dx$
$ = \frac{3}{{\sqrt 3 }}{\tan ^{ - 1}}\frac{x}{{\sqrt 3 }} - \frac{2}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 2 }}} \right) + c$
$ = \sqrt 3 {\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 3 }}} \right) - \sqrt 2 {\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 2 }}} \right) + c.$
View full question & answer→MCQ 861 Mark
$\int_{}^{} {\frac{{dx}}{{({x^2} + 1)({x^2} + 4)}} = } $
- A
$\frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{3}{\tan ^{ - 1}}\frac{x}{2} + c$
- B
$\frac{1}{3}{\tan ^{ - 1}}x + \frac{1}{3}{\tan ^{ - 1}}\frac{x}{2} + c$
- ✓
$\frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{6}{\tan ^{ - 1}}\frac{x}{2} + c$
- D
${\tan ^{ - 1}}x - 2{\tan ^{ - 1}}\frac{x}{2} + c$
AnswerCorrect option: C. $\frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{6}{\tan ^{ - 1}}\frac{x}{2} + c$
c
(c)$\int_{}^{} {\frac{{dx}}{{({x^2} + 1)({x^2} + 4)}}} = \frac{1}{3}\left[ {\int_{}^{} {\frac{{dx}}{{{x^2} + 1}} - \int_{}^{} {\frac{{dx}}{{{x^2} + 4}}} } } \right]$
$ = \frac{1}{3}\left[ {{{\tan }^{ - 1}}x - \frac{1}{2}{{\tan }^{ - 1}}\frac{x}{2}} \right] + c = \frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{6}{\tan ^{ - 1}}\frac{x}{2} + c$.
View full question & answer→MCQ 871 Mark
$\int_{}^{} {\frac{1}{{x - {x^3}}}\;dx = } $
- A
$\frac{1}{2}\log \frac{{(1 - {x^2})}}{{{x^2}}} + c$
- B
$\log \frac{{(1 - x)}}{{x(1 + x)}} + c$
- C
$\log x(1 - {x^2}) + c$
- ✓
$\frac{1}{2}\log \frac{{{x^2}}}{{(1 - {x^2})}} + c$
AnswerCorrect option: D. $\frac{1}{2}\log \frac{{{x^2}}}{{(1 - {x^2})}} + c$
d
(d)$\int_{}^{} {\frac{1}{{x - {x^3}}}\,dx = \int_{}^{} {\frac{1}{{x(1 + x)(1 - x)}}\,dx} } $
$ = \frac{1}{2}\int_{}^{} {\left( {\frac{2}{x} - \frac{1}{{1 + x}} + \frac{1}{{1 - x}}} \right)\,dx} $
$ = \frac{1}{2}[2\log x - \log (1 + x) - \log (1 - x)] = \frac{1}{2}\log \frac{{{x^2}}}{{(1 - {x^2})}} + c$.
View full question & answer→MCQ 881 Mark
$\int_{}^{} {\frac{{dx}}{{x({x^n} + 1)}} = } $
- A
$n\log \frac{{{x^n}}}{{{x^n} + 1}} + c$
- B
$n\log \frac{{{x^n} + 1}}{{{x^n}}} + c$
- ✓
$\frac{1}{n}\log \frac{{{x^n}}}{{{x^n} + 1}} + c$
- D
$\frac{1}{n}\log \frac{{{x^n} + 1}}{{{x^n}}} + c$
AnswerCorrect option: C. $\frac{1}{n}\log \frac{{{x^n}}}{{{x^n} + 1}} + c$
c
(c) Put ${x^n} = t \Rightarrow n{x^{n - 1}}dx = dt$
$ \Rightarrow \frac{{n{x^n}}}{x}\,dx = dt \Rightarrow \frac{1}{x}\,dx = \frac{{dt}}{{nt}},$ then it reduces to
$\int_{}^{} {\frac{{dt}}{{nt(t + 1)}}} = \frac{1}{n}\left[ {\int_{}^{} {\frac{{dt}}{{t(t + 1)}}} } \right]$
$ = \frac{1}{n}\left[ {\int_{}^{} {\frac{1}{t}\,dt - \int_{}^{} {\frac{1}{{t + 1}}\;dt} } } \right] = \frac{1}{n}\log \frac{{{x^n}}}{{{x^n} + 1}} + c$.
View full question & answer→MCQ 891 Mark
$\int {\frac{{x\,\,dx}}{{{x^2} + 4x + 5}} = } $
- A
$\frac{1}{2}\log ({x^2} + 4x + 5) + 2{\tan ^{ - 1}}(x) + c$
- B
$\frac{1}{2}\log ({x^2} + 4x + 5) - {\tan ^{ - 1}}(x + 2) + c$
- C
$\frac{1}{2}\log ({x^2} + 4x + 5) + {\tan ^{ - 1}}(x + 2) + c$
- ✓
$\frac{1}{2}\log ({x^2} + 4x + 5) - 2{\tan ^{ - 1}}(x + 2) + c$
AnswerCorrect option: D. $\frac{1}{2}\log ({x^2} + 4x + 5) - 2{\tan ^{ - 1}}(x + 2) + c$
d
(d)$I = \int {\frac{{x\,\,dx}}{{{x^2} + 4x + 5}}} $$ = \int {\frac{{x + 2 - 2\,\,}}{{{{(x + 2)}^2} + 1}}dx} $
$ = \frac{1}{2}\int {\frac{{2(x + 2)\,\,\,dx}}{{{{(x + 2)}^2} + 1}}} - 2\int {\frac{{dx}}{{1 + {{(x + 2)}^2}}}} $
$ = \frac{1}{2}\int {\frac{{dt}}{t} - 2\int {\frac{{dx}}{{1 + {{(x + 2)}^2}}}} } $
[Put $1 + {(x + 2)^2} = t$ in first expression ==> $2(x +2)dx = dt$]
$ = \frac{1}{2}\log t - 2{\tan ^{ - 1}}(x + 2) + c$
$ = \frac{1}{2}\log ({x^2} + 4x + 5) - 2{\tan ^{ - 1}}(x + 2) + c$.
View full question & answer→MCQ 901 Mark
$\int_{}^{} {\frac{{a\;dx}}{{b + c{e^x}}}} = $
- ✓
$\frac{a}{b}\log \left( {\frac{{{e^x}}}{{b + c{e^x}}}} \right) + c$
- B
$\frac{a}{b}\log \left( {\frac{{b + c{e^x}}}{{{e^x}}}} \right) + c$
- C
$\frac{b}{a}\log \left( {\frac{{{e^x}}}{{b + c{e^x}}}} \right) + c$
- D
$\frac{b}{a}\log \left( {\frac{{b + c{e^x}}}{{{e^x}}}} \right) + c$
AnswerCorrect option: A. $\frac{a}{b}\log \left( {\frac{{{e^x}}}{{b + c{e^x}}}} \right) + c$
a
(a)$\int_{}^{} {\frac{{a\,dx}}{{b + c\,{e^x}}} = \int_{}^{} {\frac{{a{e^x}}}{{b{e^x} + c\,{e^{2x}}}}\,dx} } $
Now put ${e^x} = t,$ then it reduces to
$a\int_{}^{} {\frac{{dt}}{{t(ct + b)}} = a\int_{}^{} { - \frac{1}{b}\left\{ {\frac{c}{{ct + b}} - \frac{1}{t}} \right\}dt} } $ {By partial fraction}
$ = \frac{a}{b}\log \left( {\frac{{{e^x}}}{{b + c{e^x}}}} \right) + c$.
View full question & answer→MCQ 911 Mark
If $\int_{}^{} {\frac{{2x + 3}}{{{x^2} - 5x + 6}}} \;dx = 9\;\ln (x - 3) - 7\ln (x - 2) + A$, then $A = $
Answerc
(c)$\int_{}^{} {\frac{{2x + 3}}{{{x^2} - 5x + 6}}dx = \int_{}^{} {\frac{{2x - 5}}{{{x^2} - 5x + 6}}dx + \int_{}^{} {\frac{8}{{{x^2} - 5x + 6}}} \;dx} } $
$ = \log [(x^2-5x+6)+ 8\int_{}^{} {\left[ {\frac{1}{{x - 2}} - \frac{1}{{x - 3}}} \right]dx + c} $.
$ = \log [(x - 2)(x - 3)] + 8\int_{}^{} {\left[ {\frac{1}{{x - 3}} - \frac{1}{{x - 2}}} \right]dx + c} $.
$ = \log (x - 2) + \log (x - 3) + 8\log (x - 3) - 8\log (x - 2) + c$
$ = 9\log (x - 3) - 7\log (x - 2) + c$.....(i)
Now given that
$\int_{}^{} {\frac{{2x + 3}}{{{x^2} - 5x + 6}}} {\rm{ }}dx = 9\log (x - 3) - 7\log (x - 2) + A$
Equating it to (i), we get $A = $constant.
View full question & answer→MCQ 921 Mark
$\int {\frac{{{x^2}}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 4} \right)}}\,} dx$ is equal to
- A
$ - {\tan ^{ - 1}}x + \frac{1}{3}{\tan ^{ - 1}}\frac{x}{2} + C$
- ✓
$- \frac{1}{3}{\tan ^{ - 1}}x + \frac{2}{3}{\tan ^{ - 1}}\frac{x}{2} + C$
- C
${\tan ^{ - 1}}x + \frac{2}{3}{\tan ^{ - 1}}\frac{x}{2} + C$
- D
$\frac{1}{3}{\tan ^{ - 1}}x - \frac{2}{3}{\tan ^{ - 1}}\frac{x}{2} + C$
AnswerCorrect option: B. $- \frac{1}{3}{\tan ^{ - 1}}x + \frac{2}{3}{\tan ^{ - 1}}\frac{x}{2} + C$
b
consider $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$ and put $x^{2}=y$
Then $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{y}{(y+1)(y+4)}$
Write $\frac{y}{(y+1)(y+4)}=\frac{A}{(y+1)}+\frac{B}{(y+4)}$
so that $y=A(y+4)+B(y+1)$
$\therefore \quad A=-\frac{1}{3} $ and $ B=\frac{4}{3}$
$\therefore \mathrm{I}=\int \frac{-\frac{1}{3}}{\left(\mathrm{x}^{2}+1\right)} \mathrm{dx}+\int \frac{4 / 3}{\left(\mathrm{x}^{2}+4\right)} \mathrm{dx}$
$=-\frac{1}{3} \tan ^{-1} \mathrm{x}+\frac{2}{3} \tan ^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{C}$
View full question & answer→MCQ 931 Mark
If $\int {f(x)\,\,dx = g(x),} $ then $\int {{f^{ - 1}}(x)} \,\,dx$ is equal to
AnswerCorrect option: B. $x{f^{ - 1}}(x) - g({f^{ - 1}}(x))$
b
(b)$\int {f(x)dx} = g(x)$
$\int {{f^{ - 1}}(x)} .1dx = {f^{ - 1}}(x)\int {dx} - \int {\left\{ {\frac{d}{{dx}}{f^{ - 1}}(x)\int {dx} } \right\}dx} $
$ = x{f^{ - 1}}(x) - \int {x\frac{d}{{dx}}{f^{ - 1}}(x)dx} $
$ = x{f^{ - 1}}(x) - \int {xd\{ {f^{ - 1}}(x)\} } $
Let ${f^{ - 1}}(x) = t$ ==> $x = f(t)$ and $d\{ {f^{ - 1}}(x)\} = dt$
$ = x{f^{ - 1}}(x) - \int {f(t)dt} = x{f^{ - 1}}(x) - g(t) = x{f^{ - 1}}(x) - g\{ {f^{ - 1}}(x)\} $.
Trick : Put $f(x) = {x^2}$, then option (b) is correct.
View full question & answer→MCQ 941 Mark
$\int_{}^{} {x{{\sec }^2}x\;dx} = $
- A
$\tan x + \log \cos x + c$
- B
$\frac{{{x^2}}}{2}{\sec ^2}x + \log \cos x + c$
- C
$x\tan x + \log \sec x + c$
- ✓
$x\tan x + \log \cos x + c$
AnswerCorrect option: D. $x\tan x + \log \cos x + c$
d
(d)$\int_{}^{} {x{{\sec }^2}x\,dx = x\tan x} - \int_{}^{} {\tan x\,dx} $
$ = x\tan x + \log (\cos x) + c.$
View full question & answer→MCQ 951 Mark
$\int_{}^{} {\left[ {\log (\log x) + \frac{1}{{{{(\log x)}^2}}}} \right]} \;dx = $
- A
$x\log (\log x) + \frac{x}{{\log x}} + c$
- ✓
$x\log (\log x) - \frac{x}{{\log x}} + c$
- C
$x\log (\log x) + \frac{{\log x}}{x} + c$
- D
$x\log (\log x) - \frac{{\log x}}{x} + c$
AnswerCorrect option: B. $x\log (\log x) - \frac{x}{{\log x}} + c$
b
(b)$\int_{}^{} {\left[ {\log (\log x) + \frac{1}{{{{(\log x)}^2}}}} \right]\,dx} = \int_{}^{} {\log (\log x)dx + \int_{}^{} {\frac{1}{{{{(\log x)}^2}}}} } dx$
$ = x\log (\log x) - \int_{}^{} {\frac{x}{{x\log x}}\,dx + \int_{}^{} {\frac{1}{{{{(\log x)}^2}}}dx} } $
$ = x\log (\log x) - \frac{x}{{\log x}} - \int_{}^{} {\frac{1}{{{{(\log x)}^2}}}dx + \int_{}^{} {\frac{1}{{{{(\log x)}^2}}}dx} } $
$ = x\log (\log x) - \frac{x}{{\log x}} + c.$
View full question & answer→MCQ 961 Mark
$\int_{}^{} {\frac{{\log x\;dx}}{{{x^3}}} = } $
- A
$\frac{1}{{4{x^2}}}(2\log x - 1) + c$
- ✓
$ - \frac{1}{{4{x^2}}}(2\log x + 1) + c$
- C
$\frac{1}{{4{x^2}}}(2\log x + 1) + c$
- D
$\frac{1}{{4{x^2}}}(1 - 2\log x) + c$
AnswerCorrect option: B. $ - \frac{1}{{4{x^2}}}(2\log x + 1) + c$
b
(b)$\int_{}^{} {\frac{{\log x}}{{{x^3}}}dx = \int_{}^{} {{x^{ - 3}}\log x\;dx} } $
$ = - \frac{{\log x}}{{2{x^2}}} + \int_{}^{} {\frac{1}{x}.\frac{1}{{2{x^2}}} + c = - \frac{{\log x}}{{2{x^2}}} + \frac{1}{2}.\frac{{{x^{ - 2}}}}{{ - 2}} + c} $
$ = - \frac{{\log x}}{{2{x^2}}} - \frac{1}{{4{x^2}}} + c = - \frac{1}{{4{x^2}}}(2\log x + 1) + c$.
View full question & answer→MCQ 971 Mark
$\int_{}^{} {x\sin x{{\sec }^3}x\,dx = } $
- A
$\frac{1}{2}[{\sec ^2}x - \tan x] + c$
- ✓
$\frac{1}{2}[x{\sec ^2}x - \tan x] + c$
- C
$\frac{1}{2}[x{\sec ^2}x + \tan x] + c$
- D
$\frac{1}{2}[{\sec ^2}x + \tan x] + c$
AnswerCorrect option: B. $\frac{1}{2}[x{\sec ^2}x - \tan x] + c$
b
(b)$\int_{}^{} {x\sin x{{\sec }^3}x\,dx} = \int_{}^{} {x\sin x\frac{1}{{{{\cos }^3}x}}\,dx} $
$ = \int_{}^{} {x\tan x\,.\,{{\sec }^2}x\,dx} $
Now put $\tan x = t \Rightarrow {\sec ^2}x\,dx = dt$ and $x = {\tan ^{ - 1}}t,$
then it reduces to $\int_{}^{} {{{\tan }^{ - 1}}t\,.\,t\,dt} = \frac{{x{{\tan }^2}x}}{2} - \frac{1}{2}t + \frac{1}{2}{\tan ^{ - 1}}t$
$ = \frac{{x({{\sec }^2}x - 1)}}{2} - \frac{1}{2}\tan x + \frac{1}{2}x = \frac{1}{2}[x{\sec ^2}x - \tan x] + c$.
View full question & answer→MCQ 981 Mark
$\int_{}^{} {{e^{2x + \log x}}} dx = $
- ✓
$\frac{1}{4}(2x - 1){e^{2x}} + c$
- B
$\frac{1}{4}(2x+ 1){e^{2x}} + c$
- C
$\frac{1}{2}(2x - 1){e^{2x}} + c$
- D
$\frac{1}{2}(2x + 1){e^{2x}} + c$
AnswerCorrect option: A. $\frac{1}{4}(2x - 1){e^{2x}} + c$
a
(a)$\int_{}^{} {{e^{2x + \log x}}dx} = \int_{}^{} {x{e^{2x}}dx} $
$ = \frac{{x{e^{2x}}}}{2} - \int_{}^{} {\frac{1}{2}{e^{2x}}dx + c} = \frac{{{e^{2x}}}}{4}(2x - 1) + c.$
View full question & answer→MCQ 991 Mark
$\int_{}^{} {\log (x + 1)dx = } $
- ✓
$(x + 1)\log (x + 1) - x + c$
- B
$(x + 1)\log (x + 1) + x + c$
- C
$(x - 1)\log (x + 1) - x + c$
- D
$(x - 1)\log (x + 1) + x + c$
AnswerCorrect option: A. $(x + 1)\log (x + 1) - x + c$
a
(a)$\int_{}^{} {\log (x + 1)\,dx} = x\log (x + 1) - \int_{}^{} {\frac{x}{{x + 1}}\,dx + c} $
$ = x\log (x + 1) - x + \log (x + 1) + c = (x + 1)\log (x + 1) - x + c$.
View full question & answer→MCQ 1001 Mark
If $\int_{}^{} {\ln ({x^2} + x)dx = x\ln ({x^2} + x) + A} $, then $A = $
Answerd
(d)$\int_{}^{} {\log ({x^2} + x)\,dx} = \int_{}^{} {\log x\,dx} + \int_{}^{} {\log (x + 1)\,dx} $
$ = x\log x - x + x\log (x + 1) - x + \log (x + 1)$
$ = x\left\{ {(\log x + \log (x + 1)} \right\} - 2x + \log (x + 1)$
$ = x\log ({x^2} + x) - 2x + \log (x + 1)$
Equating it to the given integration, we get
$A = - 2x + \log (x + 1)$.
View full question & answer→