_ ^ x^2 - 1 x^4 + x^2 + 1 dx is equal to - Vidyadip

MCQ

$\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
$\log \frac{{{x^2} - x + 1}}{{x + x + 1}} + c$

✓

Answer

Correct 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 $ $I = \frac{1}{2}\log \,\left| {\frac{{{x^2} - x + 1}}{{{x^2} + x + 1}}} \right| + c$ ==> $a = \frac{1}{2},\,b = \frac{1}{2}$.

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 "M.C.Q (1 Marks)" from STD 12 - 7.1 inderfinite integral→STD 12 - 7.1 inderfinite integral — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

The equations in terms of $x$ and $y$ are:

The value of $c$ in Rolle's theorem for the function $f(x) = x^3 - 3x$ in the interval $\big[0,\sqrt3\big]$ is:

The number of real solutions of $x^{7}+5 x^{3}+3 x+1=$ $0$ is equal to............

$\int\left(\sin ^{-1} x+\cos ^{-1} x\right) d x$ is equal to :

If $(\text{x}+2\text{y}^3)\frac{\text{dy}}{\text{dx}}=\text{y},$ then:

Let $\bar r$ be a vector in plane of $\hat i - 2\hat j + \hat k$ and $\hat i - \hat j - \hat k$ such that $\vec r.\left( {\hat i + \hat j} \right) + 2 = 0$ and length of projection of $\vec r$ on $\hat i - \hat j$ is $4\sqrt 2$, then magnitude of vector $\vec r$ is

The sum of two non$-$zero number is $8,$ the minimum value of the sum of the reciprohcle is :

If $f(x)=\left\{\begin{array}{l}2+2 x,-1 \leq x < 0 \\ 1-\frac{x}{3}, 0 \leq x \leq 3\end{array}\right.$ and $g(x)=\left\{\begin{array}{l}-x,-3 \leq x \leq 0 \\ x, 0 < x \leq 1\end{array}\right.$then range of $(fog (X))$ is

$\int_{}^{} {(x + 3){{({x^2} + 6x + 10)}^9}\;dx} $ equals

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}\,} \right| = $