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
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