Question

$\int \frac{{\left({\tan }^{-1}x\right)}^3}{1+x^2}dx=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

$I=\int \frac{{\left({\tan }^{-1}x\right)}^3}{1+x^2}dx$

Let ${\tan }^{-1}x=z$

$\Rightarrow \frac{1}{1+x^2}dx=dz$

$\therefore I=\int z^{3}dz$

$\Rightarrow I=\frac{z^4}{4}+C$

$\Rightarrow I=\frac{{\left({\tan }^{-1}x\right)}^4}{4}+C$


$\int \frac{{\left({\tan }^{-1}x\right)}^3}{1+x^2}dx=\underline{\frac{{\left({\tan }^{-1}x\right)}^4}{4}+C}.$