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Question Number 68039 by mathmax by abdo last updated on 03/Sep/19
find ∫ arctan(x+(1/x))dx
$${find}\:\int\:\:{arctan}\left({x}+\frac{\mathrm{1}}{{x}}\right){dx} \\ $$
Commented by mathmax by abdo last updated on 04/Sep/19
let I =∫ arctan(x+(1/x))dx by parts we have I =xarctan(x+(1/x))−∫ x((1−(1/x^2 ))/(1+(x+(1/x))^2 ))dx =xarctan(x+(1/x))−∫ ((x(x^2 −1))/(x^2 +(x^2 +1)^2 ))dx but J=∫ ((x(x^2 −1))/(x^2 +(x^2 +1)^2 )) =∫ ((x^3 −x)/(x^2 +x^4 +2x^2 +1))dx =∫ ((x^3 −x)/(x^4 +3x^2 +1))dx x^4 +3x^2 +1→t^2 +3t +1=0 (t=x^2 ) Δ =9−4 =5 ⇒ t_1 =((−3+(√5))/2) and t_2 =((−3−(√5))/2)) ⇒ x^4 +3x^2 +1 =t^2 +3t +1 =(t−t_1 )(t−t_2 )=(x^2 −t_1 )(x^2 −t_2 ) =(1/(t_1 −t_2 ))((1/(x^2 −t_1 ))−(1/(x^2 −t_2 ))) =(1/( (√5)))((1/(x^2 −t_1 ))−(1/(x^2 −t_2 ))) ⇒ (√5)J =∫ ((x^3 −x)/(x^2 −t_1 ))dx−∫ ((x^3 −x)/(x^2 −t_2 ))dx =∫ ((x(x^2 −t_1 )+xt_1 −x)/(x^2 −t_1 ))dx−∫ ((x(x^2 −t_2 )+xt_2 −x)/(x^2 −t_2 ))dx =(x^2 /2) +(t_1 −1) ∫ ((xdx)/(x^2 −t_1 ))−(x^2 /2)−(t_2 −1)∫ ((xdx)/(x^2 −t_2 )) +c =((t_1 −1)/2)ln∣x^2 −t_1 ∣ −((t_2 −1)/2) ln∣x^2 −t_2 ∣ +c ⇒ I =x arctan(x+(1/x))−((t_1 −1)/(2(√5)))ln∣x^2 −t_1 ∣+((t_2 −1)/(2(√5)))ln∣x^2 −t_2 ∣ +C .
$${let}\:{I}\:=\int\:{arctan}\left({x}+\frac{\mathrm{1}}{{x}}\right){dx}\:\:{by}\:{parts}\:{we}\:{have} \\ $$$${I}\:={xarctan}\left({x}+\frac{\mathrm{1}}{{x}}\right)−\int\:{x}\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\mathrm{1}+\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} }{dx} \\ $$$$={xarctan}\left({x}+\frac{\mathrm{1}}{{x}}\right)−\int\:\:\frac{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} \:+\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\:{but} \\ $$$${J}=\int\:\:\frac{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} \:+\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:=\int\:\:\frac{{x}^{\mathrm{3}} −{x}}{{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:+\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:=\int\:\:\frac{{x}^{\mathrm{3}} −{x}}{{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$${x}^{\mathrm{4}} \:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}\rightarrow{t}^{\mathrm{2}} \:+\mathrm{3}{t}\:+\mathrm{1}=\mathrm{0}\:\:\left({t}={x}^{\mathrm{2}} \right)\:\Delta\:=\mathrm{9}−\mathrm{4}\:=\mathrm{5}\:\:\Rightarrow \\ $$$$\left.{t}_{\mathrm{1}} =\frac{−\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:{and}\:{t}_{\mathrm{2}} =\frac{−\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${x}^{\mathrm{4}} \:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}\:={t}^{\mathrm{2}} \:+\mathrm{3}{t}\:+\mathrm{1}\:=\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)=\left({x}^{\mathrm{2}} −{t}_{\mathrm{1}} \right)\left({x}^{\mathrm{2}} −{t}_{\mathrm{2}} \right) \\ $$$$=\frac{\mathrm{1}}{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{t}_{\mathrm{2}} }\right)\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{t}_{\mathrm{2}} }\right)\:\Rightarrow \\ $$$$\sqrt{\mathrm{5}}{J}\:=\int\:\:\frac{{x}^{\mathrm{3}} −{x}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }{dx}−\int\:\:\frac{{x}^{\mathrm{3}} −{x}}{{x}^{\mathrm{2}} −{t}_{\mathrm{2}} }{dx} \\ $$$$=\int\:\:\frac{{x}\left({x}^{\mathrm{2}} −{t}_{\mathrm{1}} \right)+{xt}_{\mathrm{1}} −{x}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }{dx}−\int\:\:\:\frac{{x}\left({x}^{\mathrm{2}} −{t}_{\mathrm{2}} \right)+{xt}_{\mathrm{2}} −{x}}{{x}^{\mathrm{2}} \:−{t}_{\mathrm{2}} }{dx} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\left({t}_{\mathrm{1}} −\mathrm{1}\right)\:\int\:\:\frac{{xdx}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\left({t}_{\mathrm{2}} −\mathrm{1}\right)\int\:\frac{{xdx}}{{x}^{\mathrm{2}} −{t}_{\mathrm{2}} }\:+{c} \\ $$$$=\frac{{t}_{\mathrm{1}} −\mathrm{1}}{\mathrm{2}}{ln}\mid{x}^{\mathrm{2}} −{t}_{\mathrm{1}} \mid\:−\frac{{t}_{\mathrm{2}} −\mathrm{1}}{\mathrm{2}}\:{ln}\mid{x}^{\mathrm{2}} −{t}_{\mathrm{2}} \mid\:+{c}\:\Rightarrow \\ $$$${I}\:={x}\:{arctan}\left({x}+\frac{\mathrm{1}}{{x}}\right)−\frac{{t}_{\mathrm{1}} −\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}{ln}\mid{x}^{\mathrm{2}} −{t}_{\mathrm{1}} \mid+\frac{{t}_{\mathrm{2}} −\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}{ln}\mid{x}^{\mathrm{2}} −{t}_{\mathrm{2}} \mid\:+{C}\:. \\ $$$$ \\ $$$$ \\ $$

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