Question
Evaluate the following integrals:
$\int\text{x}^2\tan^{-1}\text{x dx} $

Answer

Let $\text{I}=\int\text{x}^2\tan^{-1}\text{x dx}$
$=\tan^{-1}\text{x}\int\text{x}^2\text{dx}-\int\Big(\frac{1}{1+\text{x}^2}\int\text{x}^2\text{dx}\Big)$
$=\tan^{-1}\text{x}\Big(\frac{\text{x}^3}{3}\Big)-\frac{1}{3}\frac{\text{x}^3}{1+\text{x}^2}\text{dx}$
$=\frac{1}3{\text{x}^3}\tan^{-1}\text{x}-\frac{1}{3}\int\Big(\text{x}-\frac{\text{x}}{1+\text{x}^2}\Big)\text{dx}$
$=\frac{1}{3}\text{x}^3\tan^{-1}\text{x}-\frac{1}{3}\times\frac{\text{x}^2}{2}+\frac{1}3{}\int\frac{\text{x}}{1+\text{x}^2}\text{dx}$
$\text{I}=\frac{1}{3}\text{x}^3\tan^{-1}\text{x}-\frac{1}{6}\text{x}^2+\frac{1}{6}\log\big|1+\text{x}^2\big|+\text{C}$

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