Question Number 6938 by Tawakalitu. last updated on 03/Aug/16
$${Integrate}:\:\:\:\:\:\:\:\:\frac{{x}\:{tan}^{−\mathrm{1}} \left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx} \\ $$
Commented by Yozzii last updated on 03/Aug/16
![I=∫((xtan^(−1) x)/((1+x^2 )^2 ))dx=∫(x/(1+x^2 ))tan^(−1) x×(1/(1+x^2 ))dx Let u=tan^(−1) x⇒du=(1/(1+x^2 ))dx x=tanu ∴ 1+x^2 =sec^2 u I=∫((tanu)/(sec^2 u))udu=∫usinucosudu I=∫(1/2)usin2udu I=(1/2)[(1/(−2))ucos2u−∫((−1)/2)×1×cos2udu] I=((−1)/4)ucos2u+(1/4)∫cos2udu I=((−1)/4)ucos2u+(1/8)sin2u+c tanu=x=(x/1)⇒cosu=(1/(√(1+x^2 ))), sinu=(x/(√(1+x^2 ))) cos2u=cos^2 u−sin^2 u=((1−x^2 )/(1+x^2 )) sin2u=((2x)/(1+x^2 )) ∴∫((xtan^(−1) x)/((1+x^2 )^2 ))dx=(x/(4(1+x^2 )))−(((1−x^2 )tan^(−1) x)/(4(1+x^2 )))+c](Q6940.png)
$${I}=\int\frac{{xtan}^{−\mathrm{1}} {x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\int\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{tan}^{−\mathrm{1}} {x}×\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${Let}\:{u}={tan}^{−\mathrm{1}} {x}\Rightarrow{du}=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${x}={tanu}\:\therefore\:\mathrm{1}+{x}^{\mathrm{2}} ={sec}^{\mathrm{2}} {u} \\ $$$${I}=\int\frac{{tanu}}{{sec}^{\mathrm{2}} {u}}{udu}=\int{usinucosudu} \\ $$$${I}=\int\frac{\mathrm{1}}{\mathrm{2}}{usin}\mathrm{2}{udu} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{−\mathrm{2}}{ucos}\mathrm{2}{u}−\int\frac{−\mathrm{1}}{\mathrm{2}}×\mathrm{1}×{cos}\mathrm{2}{udu}\right] \\ $$$${I}=\frac{−\mathrm{1}}{\mathrm{4}}{ucos}\mathrm{2}{u}+\frac{\mathrm{1}}{\mathrm{4}}\int{cos}\mathrm{2}{udu} \\ $$$${I}=\frac{−\mathrm{1}}{\mathrm{4}}{ucos}\mathrm{2}{u}+\frac{\mathrm{1}}{\mathrm{8}}{sin}\mathrm{2}{u}+{c} \\ $$$${tanu}={x}=\frac{{x}}{\mathrm{1}}\Rightarrow{cosu}=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }},\:{sinu}=\frac{{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${cos}\mathrm{2}{u}={cos}^{\mathrm{2}} {u}−{sin}^{\mathrm{2}} {u}=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${sin}\mathrm{2}{u}=\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\therefore\int\frac{{xtan}^{−\mathrm{1}} {x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\frac{{x}}{\mathrm{4}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}−\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right){tan}^{−\mathrm{1}} {x}}{\mathrm{4}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}+{c} \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 03/Aug/16
$${Thanks}\:{so}\:{much}..\:{i}\:{appreciate}. \\ $$
Commented by Tawakalitu. last updated on 03/Aug/16
$${Thanks}\:{fo}\:{your}\:{effort} \\ $$
Answered by Yozzii last updated on 09/Aug/16
$${An}\:{answer}\:{is}\:{in}\:{comments}. \\ $$