Question:
Write a value of $\int \frac{\left(\tan ^{-1} x\right)^{3}}{1+x^{2}} d x$
Solution:
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 $\int t^{n} d t=\frac{t^{n+1}}{n+1}$
$y=\frac{t^{4}}{4}+c$
Again, put $t=\tan ^{-1} x$
$y=\frac{\left(\tan ^{-1} x\right)^{4}}{4}+c$
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