Write a value


Write a value of $\int \frac{\left(\tan ^{-1} x\right)^{3}}{1+x^{2}} d x$


Let, $\tan ^{-1} x=t$

Differentiating both sides with respect to $x$

$\frac{d t}{d x}=\frac{1}{1+x^{2}}$

$\Rightarrow d t=\frac{d x}{1+x^{2}}$

$y=\int t^{3} d t$

Use formula $\int t^{n} d t=\frac{t^{n+1}}{n+1}$


Again, put $t=\tan ^{-1} x$

$y=\frac{\left(\tan ^{-1} x\right)^{4}}{4}+c$

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