_ ^ x^2 + 1 x^4 - x^2 + 1 ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{{x^2} + 1}}{{{x^4} - {x^2} + 1}}\;dx = } $

A
${\tan ^{ - 1}}\left( {\frac{{1 + {x^2}}}{x}} \right) + c$
B
${\cot ^{ - 1}}\left( {\frac{{1 + {x^2}}}{x}} \right) + c$
✓
${\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{x}} \right) + c$
D
${\cot ^{ - 1}}\left( {\frac{{{x^2} - 1}}{x}} \right) + c$

✓

Answer

Correct option: C.

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

c
(c)$\int_{}^{} {\frac{{{x^2} + 1}}{{{x^4} - {x^2} + 1}}\,dx = \int_{}^{} {\frac{{\left( {1 + \frac{1}{{{x^2}}}} \right)}}{{{x^2} + \frac{1}{{{x^2}}} - 1}}} } $
$ = \int_{}^{} {\frac{{1 + \frac{1}{{{x^2}}}}}{{{{\left( {x - \frac{1}{x}} \right)}^2} + 1}}\,dx = \int_{}^{} {\frac{{dt}}{{{t^2} + 1}}} = {{\tan }^{ - 1}}t + c} $
$\left\{ {{\rm{Putting}}\,x - \frac{1}{x} = t \Rightarrow \left( {1 + \frac{1}{{{x^2}}}} \right)\,dx = dt} \right\}$
$ = {\tan ^{ - 1}}\left( {x - \frac{1}{x}} \right) + c = {\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{x}} \right) + c.$

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

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Let the lines $l_1: \frac{ x +5}{3}=\frac{ y +4}{1}=\frac{ z -\alpha}{-2}$ and $l_2: 3 x +$ $2 y+z-2=0=x-3 y+2 z-13$ be coplanar. If the point $P ( a , b , c )$ on $l_1$ is nearest to the point $Q (-$ $4,-3,2)$, then $|a|+|b|+|c|$ is equal to

Let $f:R \to R$ be such that $f(1)\, = 3$ and $f'\,(1) = 6$. Then $\mathop {\lim }\limits_{x \to 0} {\left\{ {\frac{{f(1 + x)}}{{f(1)}}} \right\}^{\frac{1}{x}}}$ equals

Let $X$ be a family of sets and $R$ be a relation on $X$ defined by $‘A$ is disjoint from $B’$. Then $R$ is

If ${x_n} = \frac{{1 - 2 + 3 - 4 + 5 - 6 + ..... - 2n}}{{\sqrt {{n^2} + 1} + \sqrt {4{n^2} - 1} }},$ then $\mathop {\lim }\limits_{n \to \infty } {x_n}$ is equal to

If $e$ and $e’$ are eccentricities of hyperbola and its conjugate respectively, then

Equation of a common tangent to the circle, $x^2 + y^2 - 6x = 0$ and the parabola, $y^2 = 4x$ , is

The length of the latus rectum of a parabola, whose vertex and focus are on the positive $x$-axis at a distance $\mathrm{R}$ and $\mathrm{S}(\,>\,\mathrm{R})$ respectively from the origin, is:

On the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$ let $P$ be a point in the second quadrant such that the tangent at $\mathrm{P}$ to the ellipse is perpendicular to the line $x+2 y=0$. Let $S$ and $\mathrm{S}^{\prime}$ be the foci of the ellipse and $\mathrm{e}$ be its eccentricity. If $\mathrm{A}$ is the area of the triangle $SPS'$ then, the value of $\left(5-\mathrm{e}^{2}\right) . \mathrm{A}$ is :

${d \over {dx}}\left( {{{{{\cot }^2}x - 1} \over {{{\cot }^2}x + 1}}} \right) = $

If $y\sec x + \tan x + {x^2}y = 0$, then ${{dy} \over {dx}} =$