Mathematics

$$\int_{}^{} {\cfrac{{{x^2}{{\tan }^{ - 1}}{x^3}}}{{1 + {x^6}}}} dx$$


SOLUTION
We have,
$$I=\int_{}^{} {\cfrac{{{x^2}{{\tan }^{ - 1}}{x^3}}}{{1 + {x^6}}}} dx$$

Let
$$t=\tan^{-1}x^3$$

$$\dfrac{dt}{dx}=\dfrac{1}{1+(x^3)^2}\times 3x^2$$

$$\dfrac{dt}{3}=\dfrac{x^2}{1+x^6}dx$$

Therefore,
$$I=\dfrac{1}{3}\int t\ dt$$

$$I=\dfrac{1}{3}\dfrac{t^2}{2}+C$$

On putting the value of $$t$$, we get
$$I=\dfrac{1}{6}(\tan^{-1}x^3)^2+C$$

Hence, this is the answer.
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