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Question Number 72015 by mathmax by abdo last updated on 23/Oct/19
find ∫  ((arctan(x−(1/x)))/(x^2  +1))dx
$${find}\:\int\:\:\frac{{arctan}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$
Commented by mathmax by abdo last updated on 07/Nov/19
let A =∫  ((arctan(x−(1/x)))/(x^2  +1))dx by parts u^′ =(1/(x^2  +1))and v=arctan(x−(1/x))  A =arctanx×arctan(x−(1/x)) −∫ arctanx ×((1+(1/x^2 ))/(1+(x−(1/x))^2 ))dx  =arctan(x)×arctan(x−(1/x))−∫ arctanx×((x^2  +1)/(x^2  +(x^2 −1)^2 ))dx  ∫  (((x^2 +1)arctan(x))/(x^2  +(x^2 −1)^2 ))dx =∫  (((x^2  +1)arctan(x))/(x^2  +x^4 −2x^2  +1))dx  =∫ (((x^2  +1)arctan(x))/(x^4 −x^2  +1))dx....be continued....
$${let}\:{A}\:=\int\:\:\frac{{arctan}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:{by}\:{parts}\:{u}^{'} =\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}{and}\:{v}={arctan}\left({x}−\frac{\mathrm{1}}{{x}}\right) \\ $$$${A}\:={arctanx}×{arctan}\left({x}−\frac{\mathrm{1}}{{x}}\right)\:−\int\:{arctanx}\:×\frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\mathrm{1}+\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} }{dx} \\ $$$$={arctan}\left({x}\right)×{arctan}\left({x}−\frac{\mathrm{1}}{{x}}\right)−\int\:{arctanx}×\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{2}} \:+\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\int\:\:\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right){arctan}\left({x}\right)}{{x}^{\mathrm{2}} \:+\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dx}\:=\int\:\:\frac{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right){arctan}\left({x}\right)}{{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$=\int\:\frac{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right){arctan}\left({x}\right)}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}….{be}\:{continued}…. \\ $$

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