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Question Number 128052 by bramlexs22 last updated on 04/Jan/21
I = ∫ arctan (((x−2)/(x+2))) dx
$$\:\mathrm{I}\:=\:\int\:\mathrm{arctan}\:\left(\frac{\mathrm{x}−\mathrm{2}}{\mathrm{x}+\mathrm{2}}\right)\:\mathrm{dx}\: \\ $$
Answered by liberty last updated on 04/Jan/21
I = x arctan (((x−2)/(x+2)))−∫ x((1/(1+(((x−2)/(x+2)))^2 )). (4/((x+2)^2 )))dx I=x arctan (((x−2)/(x+2)))−∫ ((2x)/(x^2 +4)) dx I=x arctan (((x−2)/(x+2)))−ln (x^2 +4) + c
$$\mathrm{I}\:=\:\mathrm{x}\:\mathrm{arctan}\:\left(\frac{\mathrm{x}−\mathrm{2}}{\mathrm{x}+\mathrm{2}}\right)−\int\:\mathrm{x}\left(\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{x}−\mathrm{2}}{\mathrm{x}+\mathrm{2}}\right)^{\mathrm{2}} }.\:\frac{\mathrm{4}}{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} }\right)\mathrm{dx} \\ $$$$\mathrm{I}=\mathrm{x}\:\mathrm{arctan}\:\left(\frac{\mathrm{x}−\mathrm{2}}{\mathrm{x}+\mathrm{2}}\right)−\int\:\frac{\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}\:\mathrm{dx} \\ $$$$\mathrm{I}=\mathrm{x}\:\mathrm{arctan}\:\left(\frac{\mathrm{x}−\mathrm{2}}{\mathrm{x}+\mathrm{2}}\right)−\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)\:+\:\mathrm{c}\: \\ $$
Answered by Olaf last updated on 04/Jan/21
I = ∫arctan(((x−2)/(x+2)))dx I = −∫arctan(((1−(x/2))/(1+(x/2))))dx (1) I = −∫[arctan(1)−arctan((x/2))]dx (2) I = −(π/4)x+xarctan((x/2))−∫((2x)/(x^2 +4))dx I = −(π/4)x+xarctan((x/2))−ln(x^2 +4)+C (1) : artan(−u) = −arctanu (2) arctanu−arctanv = arctan(((u−v)/(1+uv)))
$$\mathrm{I}\:=\:\int\mathrm{arctan}\left(\frac{{x}−\mathrm{2}}{{x}+\mathrm{2}}\right){dx} \\ $$$$\mathrm{I}\:=\:−\int\mathrm{arctan}\left(\frac{\mathrm{1}−\frac{{x}}{\mathrm{2}}}{\mathrm{1}+\frac{{x}}{\mathrm{2}}}\right){dx}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{I}\:=\:−\int\left[\mathrm{arctan}\left(\mathrm{1}\right)−\mathrm{arctan}\left(\frac{{x}}{\mathrm{2}}\right)\right]{dx}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{I}\:=\:−\frac{\pi}{\mathrm{4}}{x}+{x}\mathrm{arctan}\left(\frac{{x}}{\mathrm{2}}\right)−\int\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{4}}{dx} \\ $$$$\mathrm{I}\:=\:−\frac{\pi}{\mathrm{4}}{x}+{x}\mathrm{arctan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{ln}\left({x}^{\mathrm{2}} +\mathrm{4}\right)+\mathrm{C} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\::\:\mathrm{artan}\left(−{u}\right)\:=\:−\mathrm{arctan}{u} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{arctan}{u}−\mathrm{arctan}{v}\:=\:\mathrm{arctan}\left(\frac{{u}−{v}}{\mathrm{1}+{uv}}\right) \\ $$
Answered by bramlexs22 last updated on 04/Jan/21

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