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
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