Mathematics

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

##### SOLUTION

Let $I=\displaystyle\int_{0}^{1}\dfrac{\tan^{-1}x}{1+x^{2}}dx$

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

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

Also

$x=0\Rightarrow t=\tan^{-1}0=0$

$x=1\Rightarrow t=\tan^{-1}1=\dfrac{\pi}{4}$

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

$=\left[\dfrac{t^{2}}{2}\right]_{0}^{\pi/4}$  [$\because \int x^n=\dfrac{x^{n+1}}{n+1}$]

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

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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