Mathematics

# Evaluate $\displaystyle\int_{0}^{1}\dfrac{\tan^{-1}x}{1+x^{2}}dx$

##### SOLUTION
Consider, $I=\displaystyle\int_{0}^{1}\dfrac{\tan^{-1}x}{1+x^{2}}dx$

Let, $\tan^{-1}x=t$.

Then, $d(\tan^{-1}x)=dt\Rightarrow dx=(1+x^{2})dt$

Also, $x=0\Rightarrow t=\tan^{-1}0=0$ and $x=1\Rightarrow t=\tan^{-1}1=\dfrac{\pi}{4}$

$\therefore I=\displaystyle\int_{0}^{\pi/4}t\ dt$

$I=\left[\dfrac{t^{2}}{2}\right]_{0}^{\pi/4}$

$I=\dfrac{\pi^{2}}{32}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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