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Question Number 62937 by Prithwish sen last updated on 27/Jun/19

$${\int }_0^{\mathrm{x}}\frac{1}{1+{\mathrm{x}}^2}\mathrm{dx}$$

Commented by mathmax by abdo last updated on 27/Jun/19

$${\int }_0^x\frac{dt}{1+t^2}={\left[arctan\left(t\right)\right]}_0^x=arctanx.$$

Commented by Prithwish sen last updated on 27/Jun/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by Prithwish sen last updated on 27/Jun/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by peter frank last updated on 27/Jun/19

$$letx=\tan \theta \to \theta ={\tan }^{-1}x$$$$dx=\sec {}^2\theta d\theta$$$$\int \frac{\sec {}^2\theta d\theta }{\sec {}^2\theta }$$$$\theta +c$$$${\tan }^{-1}x+c$$$$$$

Commented by Prithwish sen last updated on 27/Jun/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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