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Question Number 61614 by Tajaddin last updated on 05/Jun/19

$$\underset{-1}{\stackrel{1}{\int }}\frac{d}{dx}\left({\tan }^{-1}\frac{1}{x}\right)dx=$$

Commented by maxmathsup by imad last updated on 05/Jun/19

$$wehave\frac{d}{dx}\left(arctan\left(\frac{1}{x}\right)\right)=\frac{-1}{x^2(1+\frac{1}{x^2})}=\frac{-1}{x^2+1}\Rightarrow$$$${\int }_{-1}^1\frac{d}{dx}\left(arctan\left(\frac{1}{x}\right)\right)dx=-{\int }_{-1}^1\frac{dx}{1+x^2}=-2{\int }_0^1\frac{dx}{1+x^2}=-2{\left[arctanx\right]}_0^1=-2\frac{\pi }{4}=-\frac{\pi }{2}$$

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