Integration Questions

Question Number 123526 by bramlexs22 last updated on 26/Nov/20

$$\:\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{x}\:\mathrm{arctan}\:{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:? \\$$

Commented by liberty last updated on 26/Nov/20

$$\frac{\pi}{\mathrm{8}}\:? \\$$

Commented by bramlexs22 last updated on 26/Nov/20

$${how}\:{sir} \\$$

Answered by Lordose last updated on 26/Nov/20

$$\\$$$$\mathrm{I}\:=\:\int_{\:\mathrm{0}} ^{\:\infty} \frac{\mathrm{xtan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx} \\$$$$\mathrm{IBP} \\$$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx}\:−\:\mid\frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{2}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mid_{\mathrm{0}} ^{\infty} \\$$$$\mathrm{Ostrogradski}\:\mathrm{method} \\$$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left[\left(\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)\right]_{\mathrm{0}} ^{\infty} −\:\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\right]_{\mathrm{0}} ^{\infty} \\$$$$\mathrm{I}\:=\:\frac{\pi}{\mathrm{8}} \\$$

Commented by bramlexs22 last updated on 26/Nov/20

$${thank}\:{you} \\$$

Answered by Dwaipayan Shikari last updated on 26/Nov/20

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {t}\:{sint}\:{cost}\:{dt}\:\:\:\:\:\:\:\:\:\:\:\:\:{t}={tan}^{−\mathrm{1}} {x} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {tsin}\mathrm{2}{t}\:{dt}\:=−\left[\frac{{t}}{\mathrm{4}}{cos}\mathrm{2}{t}\right]^{\frac{\pi}{\mathrm{2}}} \:+\frac{\mathrm{1}}{\mathrm{8}}\left[{sin}\mathrm{2}{t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{dt} \\$$$$=\frac{\pi}{\mathrm{8}} \\$$

Commented by bramlexs22 last updated on 26/Nov/20

$${thank}\:{you} \\$$

Answered by mathmax by abdo last updated on 26/Nov/20

$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{xarctanx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx}\:\mathrm{we}\:\mathrm{integrate}\:\mathrm{by}\:\mathrm{parts}\:\mathrm{u}^{'} \:=\frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\$$$$\mathrm{and}\:\mathrm{v}=\mathrm{srctsnx}\:\Rightarrow\mathrm{A}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{arctanx}\right]_{\mathrm{0}} ^{\infty} +\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{x}^{\left.\mathrm{2}\right)} \right.}\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=_{\mathrm{x}=\mathrm{tan}\theta} \:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta}{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\mathrm{d}\theta \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} \theta\:\mathrm{d}\theta\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+\mathrm{cos}\left(\mathrm{2}\theta\right)\right)\mathrm{d}\theta \\$$$$=\frac{\pi}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left[\mathrm{sin}\left(\mathrm{2}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\frac{\pi}{\mathrm{8}}+\mathrm{0}\:=\frac{\pi}{\mathrm{8}} \\$$$$\mathrm{A}\:=\frac{\pi}{\mathrm{8}} \\$$