$\int \frac{\left(\tan ^{-1} x\right)^3}{1+x^2} d x$ is equal to
Answer & explanation
Correct answer: option 2
We have,
$\int \frac{\left(\tan ^{-1} x\right)^3}{1+x^2} d x=\int\left(\tan ^{-1} x\right)^3 d\left(\tan ^{-1} x\right)=\frac{\left(\tan ^{-1} x\right)^4}{4}+C$