_ ^ 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

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

Let $x , y , z > 1$ and $A=\left[\begin{array}{lll}1 & \log _x y & \log _x z \\ \log _y x & 2 & \log _y z \\ \log _z x & \log _z y & 3\end{array}\right]$ .Then $\left|\operatorname{adj}\left(\operatorname{adj} A^2\right)\right|$ is equal to

If $A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$and $B = \left[ {\begin{array}{*{20}{c}}{\cos \beta }&{ - \sin \beta }\\{\sin \beta }&{\cos \beta }\end{array}} \right]$, then the correct relation is

Solve the system of equations $ 2 x+5 y=1 \,;\,3 x+2 y=7$

The solution of the differential equation ${x^4}\frac{{dy}}{{dx}} + {x^3}y + {\rm{cosec}}\,(xy) = 0$ is equal to

A unit vector in the plane of $i + 2j + k$ and $i + j + 2k$are perpendicular to $2i + j + k$ is

Let a unit vector which makes an angle of $60^{\circ}$ with $2 \hat{i}+2 \hat{j}-\hat{k}$ and an angle of $45^{\circ}$ with $\hat{i}-\hat{k}$ be $\overrightarrow{\mathrm{C}}$. Then $\overrightarrow{\mathrm{C}}+\left(-\frac{1}{2} \hat{\mathrm{i}}+\frac{1}{3 \sqrt{2}} \hat{\mathrm{j}}-\frac{\sqrt{2}}{3} \hat{\mathrm{k}}\right)$ is :

ABCD is a parallelogram with AC and BD as diagonals. Then, $\overrightarrow{\text{AC}}-\overrightarrow{\text{BD}}=$

Let $S$ be the set of all real values of $k$ for which the system oflinear equations $x +y + z = 2$ ; $2x +y - z = 3$ ; $3x + 2y + kz = 4$ has a unique solution. Then $S$ is

A(3, 2, 0), B(5, 3, 2) and C(-9, 6, -3) are the vertices of a tringle ABC. if the bisector of $\angle\text{ABC}$ meets BC at D, then coordinates of D are:

Choose the correct answer from given four options in each of the Exercise:The maximum value of $\Delta=\begin{vmatrix}1 & 1&1 \\1 &1+\sin\theta&1\\1+\cos\theta &1&1\end{vmatrix}$ is $(\theta$ is real number$):$