Evaluate the following integrals: _ 0 ^ 1 x ^ -1 x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\limits_{0}^{1}\text{x}\tan^{-1}\text{x}\text{ dx} $

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Answer

We have,
$\int_{0}^\limits{1}\text{x}\tan^{-1}\text{x}\text{ dx}=\tan^{-1}\text{x}\int_{0}^\limits{1}\text{x dx}-\int_{0}^\limits{1}\big(\int\text{dx}\big)\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{x}\big)\text{dx}$
$=\Big[\frac{\text{x}^2}{2}\tan^{-1}\text{x}\Big]^1_0-\frac{1}{2}\int_{0}^\limits{1}\frac{\text{x}^2}{1+\text{x}^2}\text{ dx}$
$=\Big[\frac{\text{x}^2}{2}\tan^{-1}\text{x}\Big]^1_0-\frac{1}{2}\int_{0}^\limits{1}\frac{1+\text{x}^2-1}{1+\text{x}^2}\text{ dx}$
$=\frac{1}{2}\Big(\frac{\pi}{4}\Big)-\frac{1}{2}\Bigg[\int_{0}^\limits{1}\text{dx}-\int_{0}^\limits{1}\frac{\text{dx}}{1+\text{x}^2}\Bigg]$
$=\frac{\pi}{8}-\frac{1}{2}\big[\text{x}-\tan^{-1}\text{x}\big]^{1}_0$
$=\frac{\pi}{8}-\frac{1}{2}\Big[1-\frac{\pi}{4}\Big]$
$=\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}$
$=\frac{\pi}{4}-\frac{1}{2}$

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