_ ^ 1 ( x^2 + a^2 )( x^2 + b^2 ) dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{({x^2} + {a^2})({x^2} + {b^2})}}dx = } $

✓
$\frac{1}{{({a^2} - {b^2})}}\left[ {\frac{1}{b}{{\tan }^{ - 1}}\left( {\frac{x}{b}} \right) - \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)} \right] + c$
B
$\frac{1}{{({b^2} - {a^2})}}\left[ {\frac{1}{b}{{\tan }^{ - 1}}\left( {\frac{x}{b}} \right) - \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)} \right] + c$
C
$\frac{1}{b}{\tan ^{ - 1}}\left( {\frac{x}{b}} \right) - \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + c$
D
$\frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) - \frac{1}{b}{\tan ^{ - 1}}\left( {\frac{x}{b}} \right) + c$

✓

Answer

Correct option: A.

$\frac{1}{{({a^2} - {b^2})}}\left[ {\frac{1}{b}{{\tan }^{ - 1}}\left( {\frac{x}{b}} \right) - \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)} \right] + c$

(a)$\int_{}^{} {\frac{1}{{({x^2} + {b^2})({x^2} + {a^2})}}} \,dx$
$ = \frac{1}{{{a^2} - {b^2}}}\int_{}^{} {\left[ {\frac{1}{{{x^2} + {b^2}}} - \frac{1}{{{x^2} + {a^2}}}} \right]} \,dx$
$ = \frac{1}{{({a^2} - {b^2})}}\left[ {\frac{1}{b}{{\tan }^{ - 1}}\left( {\frac{x}{b}} \right) - \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)} \right] + c$.
Note : Students should remember this question as a formula.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from 7.1 inderfinite integral→7.1 inderfinite integral — all question types→ગણિત ધોરણ 12 સાયન્સ question papers→Gujarat Board ધોરણ 12 સાયન્સ question papers→Gujarat Board question paper generator→

Similar questions

Let $E^c$ denote the complement of an event $E$. Let $E, F, G$ be pairwise independent events with $P(G)>0$ and $P(E \cap F \cap G)$ $=0$. Then $P\left(E^C \cap F^C \mid G\right)$ equals

એક જ ઊગમબિંદુવાળા લંબવૃત્તીય અક્ષોની બે યામ ૫દ્ઘતિ છે. એક સમતલ ઊગમબિંદુથી પ્રથમ યામ ૫દ્ઘતિના અક્ષોને $a,b,c$ અંતરે અને બીજી યામ ૫દ્ઘતિના અક્ષોને $a ',b ',c '$ અંત૨ે છેદે , તો $........ .$

વક્ર $v = \frac{A}{r} + B $ નું વિકલ સમીકરણ મેળવો. (કે જ્યાં $A$ અને $B$એ સ્વૈર અચળાંક છે )

$\int\limits_{ - 2}^0 {\left\{ {{x^3} + 3{x^2} + 3x + 3 + \left( {x + 1} \right)\cos \left( {x + 1} \right)} \right\}\,\,dx =\ .....} $

Suppose thatBox$-I$ contains $8$ red, $3$ blue and $5$ green balls,
Box$-II$ contains $24$ red, $9$ blue and $15$ green balls,
Box$-III$ contains $1$ blue, $12$ green and $3$ yellow balls,
Box$-IV$ contains $10$ green, $16$ orange and $6$ white balls.
A ball is chosen randomly from Box-I ; call this ball $b$. If $b$ is red then a ball is chosen randomly from Box$-II$, if $b$ is blue then a ball is chosen randomly from Box$-III$, and if $b$ is green then a ball is chosen randomly from Box$-IV$. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened, is equal to

જો $y = {\log _2}[{\log _2}(x)]$, તો ${{dy} \over {dx}}= . . . .$

વિકલ સમીકરણ $\frac{{dy}}{{dx}} = \frac{{xy}}{{{x^2} + {y^2}}}$ નો ઉકેલ મેળવો.

$\int_{}^{} {\frac{1}{{{{\cos }^{ - 1}}x.\sqrt {1 - {x^2}} }}dx = } $

$\int_0^\pi {\frac{{dx}}{{1 + \sin x}}} = $

જો વક્ર $2{y^2} = a{x^2} + {x^2}$ નો બિંદુ $\left( {a,a} \right)$આગળનો સ્પર્શક અક્ષો પર $\alpha $ તથા $\beta $ અંતઃખંડો કાપતા હોય તથા જો ${\alpha ^2} + {\beta ^2} = 61$ હોય તો $|a| =\ ..........$