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$$
\int \frac{4 x^2 \cot ^{-1}\left(x^3\right)}{1+x^6} \mathrm{~d} x=
$$
(where $C$ is a constant of integration.)
Options:
\int \frac{4 x^2 \cot ^{-1}\left(x^3\right)}{1+x^6} \mathrm{~d} x=
$$
(where $C$ is a constant of integration.)
Solution:
2962 Upvotes
Verified Answer
The correct answer is:
$\frac{-2}{3}\left(\cot ^{-1} X^3\right)^2+C$
$\int \frac{4 x^2 \cot ^{-1}\left(x^3\right)}{1+x^6} \mathrm{~d} x$ let $\cot ^{-1}\left(x^3\right)=t$
then $\frac{-3 x^2}{1+x^6} d x=d t$
$\begin{aligned} & =-\frac{4}{3} \int t \mathrm{~d} t \\ & =-\frac{4}{3} \times \frac{t^2}{2}+c=-\frac{2}{3}\left(\cot ^{-1} X^3\right)^2+c\end{aligned}$
then $\frac{-3 x^2}{1+x^6} d x=d t$
$\begin{aligned} & =-\frac{4}{3} \int t \mathrm{~d} t \\ & =-\frac{4}{3} \times \frac{t^2}{2}+c=-\frac{2}{3}\left(\cot ^{-1} X^3\right)^2+c\end{aligned}$
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