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Question Number 93473 by mashallah last updated on 13/May/20

$$\int 1/{(1+{\mathrm{x}}^2)}^2$$

Commented by abdomathmax last updated on 15/May/20

$$A=\int \frac{dx}{{(1+x^2)}^2}wedothechangementx=tant\Rightarrow$$$$A=\int \frac{(1+{tan}^{2}t)dt}{{(1+{tan}^{2}t)}^2}=\int {cos}^{2}tdt=\frac{1}{2}\int (1+cos\left(2t\right))dt$$$$=\frac{t}{2}+\frac{1}{4}sin\left(2t\right)+C$$$$=\frac{arctanx}{2}+\frac{1}{4}\times \frac{tant}{1+{tan}^{2}t}+C$$$$=\frac{arctanx}{2}+\frac{1}{4}\frac{x}{1+x^2}+C$$$$$$

Answered by Rio Michael last updated on 13/May/20

$$\int \frac{1}{{(1+x^2)}^2}dx$$$$\mathrm{changement}x=\tan \theta \Rightarrow \frac{dx}{d\theta }={\sec }^2\theta$$$$\Rightarrow \int \frac{1}{{(1+x^2)}^2}dx=\int \frac{1}{{(1+{\tan }^2\theta )}^2}{\sec }^2\theta d\theta =\int \frac{1}{{\sec }^2\theta }d\theta$$$$\Rightarrow \int {\cos }^2\theta d\theta =\frac{1}{2}\int (1+\cos 2\theta )d\theta =\frac{1}{2}(\theta +\frac{\sin 2\theta }{2})+C$$$$$$

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