Evaluate the following integrals: ( ^ -1 x^2)x dx - Vidyadip

Question

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

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Answer

Let $\text{I}=\int(\tan^{-1}\text{x}^2)\text{x dx}$
Let $\text{x}^2=\text{t}$
$2\text{x dx = dt}$
$\text{I}=\frac{1}{2}\int\tan^{-1}\text{t dt}$
$=\frac{1}{2}\int1\tan^{-1}\text{t dt}$
$=\frac{1}{2}\Big[\tan^{-1}\text{t}\int\text{dt}-\Big(\int\frac{1}{1+\text{t}^2}\int\text{dt}\Big)\text{dt}\Big]$
$=\frac{1}{2}\Big[\text{t}\tan^{-1}\text{t}-\int\frac{\text{t}}{1+\text{t}^2}\text{dt}\Big]$
$=\frac{1}{2}\text{t}\tan^{-1}\text{t}-\frac{1}{4}\int\frac{2\text{t}}{1+\text{t}^2}\text{dt}$
$=\frac{1}{2}\text{t}\tan^{-1}\text{t}-\frac{1}{4}\log\big|1+\text{t}^2\big|+\text{C}$
$\text{I}=\frac{1}{2}\text{x}^2\tan^{-1}\text{x}^2-\frac{1}{4}\log\big|1+\text{x}^4\big|+\text{C}$

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