Question

If $\int \frac{x^2{\tan }^{-1}{x}^3}{1+x^6}dx=\frac{p}{q}\left(f\right(x){)}^r+C,p,q$ are co-prime numbers, $r\in I,$ then the value of $\frac{pqrf\left(1\right)}{\pi }$ is
(where $C$ is integration constant)

Answer 1

put $\left({\tan }^{-1}{x}^3\right)=z$
$\Rightarrow \frac{1}{1+x^6}\cdot 3x^{2}dx=dz$
Now,
$I=\int \frac{x^2\left({\tan }^{-1}{x}^3\right)dx}{1+x^6}=\int \frac{1}{3}zdz=\frac{1}{6}{z}^2+C$
$=\frac{1}{6}({\tan }^{-1}{x}^3{)}^2+C$
$\therefore p=1,q=6,r=2,f\left(x\right)=\left({\tan }^{-1}{x}^3\right)\Rightarrow f\left(1\right)=\frac{\pi }{4}$
$\Rightarrow \frac{pqrf\left(1\right)}{\pi }=1\times 6\times 2\times \frac{\pi }{4}\times \frac{1}{\pi }=3$