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Question Number 40133 by maxmathsup by imad last updated on 16/Jul/18

$$find{\int }_0^1\frac{dt}{{(1+t^2)}^2}$$

Commented by maxmathsup by imad last updated on 18/Jul/18

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

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