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Question Number 36431 by prof Abdo imad last updated on 02/Jun/18

$$calculate{\int }_0^1\frac{x+1}{{(x^2+1)}^2}dx$$

Commented by prof Abdo imad last updated on 03/Jun/18

$$I=\frac{1}{2}{\int }_0^1\frac{2x+2}{{(x^2+1)}^2}dx=\frac{1}{2}\int \frac{2x}{{(x^2+1)}^2}+{\int }_0^1\frac{dx}{{(x^2+1)}^2}$$$$={\left[\frac{-1}{2(x^2+1)}\right]}_0^1+{\int }_0^1\frac{dx}{{(x^2+1)}^2}changement$$$$x=tan\theta give$$$${\int }_0^1\frac{dx}{{(1+x^2)}^2}={\int }_0^{\frac{\pi }{4}}\frac{(1+{tan}^2\theta )}{{(1+{tan}^2\theta )}^2}d\theta$$$$={\int }_0^{\frac{\pi }{4}}{cos}^2\theta d\theta ={\int }_0^{\frac{\pi }{4}}\frac{1+cos\left(2\theta \right)}{2}d\theta$$$$=\frac{\pi }{8}+\frac{1}{4}{\left[sin\left(2\theta \right)\right]}_0^{\frac{\pi }{4}}=\frac{\pi }{8}+\frac{1}{4}\Rightarrow$$$$I=\frac{1}{2}-\frac{1}{4}+\frac{\pi }{8}+\frac{1}{4}\Rightarrow$$$$I=\frac{\pi }{8}+\frac{1}{2}.$$

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