Question

$\int \frac{{\left({\tan }^{-1}x\right)}^3}{1+x^2}dx$ is equal to

• A. $3{\left({\tan }^{-1}x\right)}^2$
• B. $\frac{{\left({\tan }^{-1}x\right)}^4}{4}+c$
• C. ${\left({\tan }^{-1}x\right)}^4+c$
• D. None of these

Answer 1

Answer: (B)

Let $I=\int \frac{{\left({\tan }^{-1}x\right)}^3}{1+x^2}dx$. Put ${\tan }^{-1}x=t$
$\Rightarrow \frac{1}{1+x^2}dx=dt$
$\therefore I=\int t^{3}dt=\frac{t^4}{4}+c$
$=\frac{{\left({\tan }^{-1}x\right)}^4}{4}+c$