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

∫_(−1) ^1  (d/dx) (tan^(−1) (1/x))dx =

$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\:\frac{{d}}{{dx}}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}\right){dx}\:= \\ $$

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

we have (d/dx)(arctan((1/x))) =((−1)/(x^2 (1+(1/x^2 )))) =((−1)/(x^2  +1)) ⇒  ∫_(−1) ^1  (d/dx)(arctan((1/x)))dx =−∫_(−1) ^1   (dx/(1+x^2 )) =−2 ∫_0 ^1  (dx/(1+x^2 )) =−2[arctanx]_0 ^1 =−2(π/4) =−(π/2)

$${we}\:{have}\:\frac{{d}}{{dx}}\left({arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)\:=\frac{−\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\:=\frac{−\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{{d}}{{dx}}\left({arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right){dx}\:=−\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=−\mathrm{2}\left[{arctanx}\right]_{\mathrm{0}} ^{\mathrm{1}} =−\mathrm{2}\frac{\pi}{\mathrm{4}}\:=−\frac{\pi}{\mathrm{2}} \\ $$

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