$\int \frac{4 x^2 \cot ^{-1}\left(x^3\right)}{1+x^6} \mathrm{~d} x=$ (where $C$ is a constant of integration.)

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Aicharya · Accepted Answer

$\int \frac{4 x^2 \cot ^{-1}\left(x^3ight)}{1+x^6} \mathrm{~d} x$ let $\cot ^{-1}\left(x^3ight)=t$ then $\frac{-3 x^2}{1+x^6} d x=d t$ $\begin{aligned} & =-\frac{4}{3} \int t \mathrm{~d} t \ & =-\frac{4}{3} 	imes \frac{t^2}{2}+c=-\frac{2}{3}\left(\cot ^{-1} X^3ight)^2+c\end{aligned}$

$\int \frac{4 x^2 \cot ^{-1}\left(x^3\right)}{1+x^6} \mathrm{~d} x=$ (where $C$ is a constant of integration.)

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