_ ^ x^2 ^ - 1 x^3 1 + x^6 ;dx is equal to - Vidyadip

MCQ

$\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
$\frac{1}{2}{({\tan ^{ - 1}}{x^2})^3} + c$

✓

Answer

Correct 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$.

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 7.1 inderfinite integral→7.1 inderfinite integral — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

$\int\limits^\pi_0\frac{1}{\text{a}+\text{b}\cos\text{x}}\text{ dx}=$

$\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$
$\frac{\pi}{\text{ab}}$
$\frac{\pi}{\text{a}^2-\text{b}^2}$
$({\text{a}+\text{b}})\pi$

The area bounded by the parabola $x^2=4 y$ and its latus rectum is:

If $\vec{\text{a}}$ and $\vec{\text{b}}$ be two unit vectors and $\theta$ the angle between them, than $\vec{\text{a}}+\vec{\text{b}}$ is a unit vector if $\theta=$

$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\frac{\pi}{2}$
$\frac{2\pi}{3}$

The value of the integral $\int_{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x$, where $a > 0$, is

If log_e 4 = 1.3868, then log_e 4.01 =

1.3968
1.3898
1.3893
None of these

Let $\mathrm{A}$ and $\mathrm{B}$ be two square matrices of order $3$ such that $|A|=3$ and $|B|=2$. Then $\left|\mathrm{A}^{\mathrm{T}} \mathrm{A}(\operatorname{adj}(2 \mathrm{~A}))^{-1}(\operatorname{adj}(4 \mathrm{~B}))(\operatorname{adj}(\mathrm{AB}))^{-1} \mathrm{AA}^{\mathrm{T}}\right|$ is equal to :

If $\Delta=\left|\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|$ and $\mathrm{A}_{i j}$ is Cofactors of $a_{i j},$ then value of $\Delta$ is given by

Let $a_1,a_2,a_3,....,a_{10}$ be in $G.P.$ with $a_i > 0$ for $i = 1, 2,....,10$ and $S$ be the set of pairs $(r,k), r, k \in N$ (the set of natural numbers) for which

$\left| {\begin{array}{*{20}{c}}
{{{\log }_e}\,a_1^ra_2^k}&{{{\log }_e}\,a_2^ra_3^k}&{{{\log }_e}\,a_3^ra_4^k} \\
{{{\log }_e}\,a_4^ra_5^k}&{{{\log }_e}\,a_5^ra_6^k}&{{{\log }_e}\,a_6^ra_7^k} \\
{{{\log }_e}\,a_7^ra_8^k}&{{{\log }_e}\,a_8^ra_9^k}&{{{\log }_e}\,a_9^ra_{10}^k}
\end{array}} \right| = 0$

Then the number of elements in $S$, is

If $A$ is a square matrix of order 3 and $|A|=5$, then $|\operatorname{adj} A|=$

For the following LPP, maximise $Z=3 x+4 y$ subject to constraints $x-y \geq-1, x \leq 3, x \geq 0, y \geq 0$ the maximum value is