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Question Number 92138 by hore last updated on 05/May/20

$$\int \frac{dx}{1+x^2}$$

Answered by john santu last updated on 05/May/20

$${\tan }^{-1}\left(\mathrm{x}\right)+\mathrm{c}$$

Answered by Rio Michael last updated on 07/May/20

$$\int \frac{dx}{1+x^2}$$$$\mathrm{let}x=\tan \theta \Rightarrow \frac{dx}{d\theta }={\sec }^2\theta$$$$\Rightarrow \int \frac{dx}{1+x^2}=\int \frac{1}{1+{\tan }^2\theta }{\sec }^2\theta d\theta$$$$=\int \frac{{\sec }^2\theta }{{\sec }^2\theta }d\theta {}_{1+{\tan }^2\theta ={\sec }^2\theta }$$$$=\int d\theta =\theta +C$$$$\mathrm{but}x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x$$$$\Rightarrow \int \frac{dx}{1+x^2}={\tan }^{-1}x+C$$

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