_ ^ 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$.

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