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Question Number 40870 by math khazana by abdo last updated on 28/Jul/18

fnd ∫ (1+(1/x^2 ))arctan(x−(1/x))dx .

$${fnd}\:\:\int\:\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctan}\left({x}−\frac{\mathrm{1}}{{x}}\right){dx}\:. \\ $$

Commented by maxmathsup by imad last updated on 29/Jul/18

changement x−(1/x)=t give (1+(1/x^2 ))dx=dt ⇒ I = ∫ arctant dt = t arctant −∫ (t/(1+t^2 ))dt +c =t arctan(t)−(1/2)ln(1+t^2 )+c =(x−(1/x))arctan(x−(1/x))−(1/2)ln(1+(x−(1/x))^2 ) +c

$${changement}\:{x}−\frac{\mathrm{1}}{{x}}={t}\:{give}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx}={dt}\:\Rightarrow \\ $$$${I}\:=\:\int\:\:\:\:{arctant}\:{dt}\:=\:{t}\:{arctant}\:−\int\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:+{c} \\ $$$$={t}\:{arctan}\left({t}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)+{c} \\ $$$$=\left({x}−\frac{\mathrm{1}}{{x}}\right){arctan}\left({x}−\frac{\mathrm{1}}{{x}}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \right)\:+{c} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18

t=x−(1/x) dt=1+(1/x^(2 ) ) dt ∫tan^(−1) dt t ×tan^(−1) t−∫(1/(1+t^2 ))×tdt t×tan^(−1) t−(1/2)ln(1+t^2 )+c (x−(1/x))tan^(−1) (x−(1/x))−(1/2)ln∣(1+x^2 +(1/x^2 )−2)∣+c (x−(1/x))tan^(−1) (x−(1/x))−(1/2)ln∣x^2 +(1/x^2 )−1∣+c

$${t}={x}−\frac{\mathrm{1}}{{x}}\:\:\:{dt}=\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}\:} }\:{dt} \\ $$$$\int{tan}^{−\mathrm{1}} {dt} \\ $$$${t}\:×{tan}^{−\mathrm{1}} {t}−\int\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }×{tdt} \\ $$$${t}×{tan}^{−\mathrm{1}} {t}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)+{c} \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}\right){tan}^{−\mathrm{1}} \left({x}−\frac{\mathrm{1}}{{x}}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\left(\mathrm{1}+{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{2}\right)\mid+{c} \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}\right){tan}^{−\mathrm{1}} \left({x}−\frac{\mathrm{1}}{{x}}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}\mid+{c} \\ $$

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