Question
$\int x^3 \tan ^{-1} x d x$

Answer

$\text { Let } I =\int x^3 \cdot \tan ^{-1} x \cdot d x$
$=\int\left(\tan ^{-1} x\right) x^3 d x$
$=\left(\tan ^{-1} x\right) \int x^3 d x-\int\left[\frac{ d }{ d x}\left(\tan ^{-1} x\right) \int x^3 d x\right] d x$
$=\left(\tan ^{-1} x\right) \cdot\left(\frac{x^4}{4}\right)-\int \frac{1}{1+x^2} \cdot \frac{x^4}{4} d x$
$=\frac{x^4}{4} \tan ^{-1} x+\frac{1}{4} \int \frac{\left(-x^4\right)}{1+x^2} \cdot d x$
$=\frac{x^4}{4} \tan ^{-1} x+\frac{1}{4} \int \frac{\left(1-x^4\right)-1}{1+x} d x$
$=\frac{x^4}{4} \tan ^{-1} x+\frac{1}{4} \int \frac{\left(1-x^2\right)\left(1+x^2\right)-1}{1+x^2} d x$
$=\frac{x^4}{4} \tan ^{-1} x+\frac{1}{4} \int\left(1-x^2-\frac{1}{1+x^2}\right) d x$
$=\frac{x^4}{4} \tan ^{-1} x+\frac{1}{4}\left(x-\frac{x^3}{3}-\tan ^{-1} x\right)+ c $

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