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

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

Commented by liberty last updated on 26/Nov/20

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

Commented by bramlexs22 last updated on 26/Nov/20

$$howsir$$

Answered by Lordose last updated on 26/Nov/20

$$$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{xtan}}^{-1}\left(\mathrm{x}\right)}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}$$$$\mathrm{IBP}$$$$\mathrm{I}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}-\mid \frac{{\tan }^{-1}\left(\mathrm{x}\right)}{2(1+{\mathrm{x}}^2)}{\mid }_0^{\mathrm{\infty }}$$$$\mathrm{Ostrogradski}\mathrm{method}$$$$\mathrm{I}=\frac{1}{4}{\left[(\frac{\mathrm{x}}{1+{\mathrm{x}}^2}+{\tan }^{-1}\left(\mathrm{x}\right))\right]}_0^{\mathrm{\infty }}-\frac{1}{2}{\left[\frac{{\tan }^{-1}\left(\mathrm{x}\right)}{(1+{\mathrm{x}}^2)}\right]}_0^{\mathrm{\infty }}$$$$\mathrm{I}=\frac{\pi }{8}$$

Commented by bramlexs22 last updated on 26/Nov/20

$$thankyou$$

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

$${\int }_0^{\frac{\pi }{2}}tsintcostdtt={tan}^{-1}x$$$$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}tsin2tdt=-{\left[\frac{t}{4}cos2t\right]}^{\frac{\pi }{2}}+\frac{1}{8}{\left[sin2t\right]}_0^{\frac{\pi }{2}}dt$$$$=\frac{\pi }{8}$$

Commented by bramlexs22 last updated on 26/Nov/20

$$thankyou$$

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

$$\mathrm{A}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{xarctanx}}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}\mathrm{we}\mathrm{integrate}\mathrm{by}\mathrm{parts}{\mathrm{u}}^{{}^{\prime }}=\frac{\mathrm{x}}{{(1+{\mathrm{x}}^2)}^2}$$$$\mathrm{and}\mathrm{v}=\mathrm{srctsnx}\Rightarrow \mathrm{A}={[-\frac{1}{2(1+{\mathrm{x}}^2)}\mathrm{arctanx}]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\frac{1}{2(1+{\mathrm{x}}^{2)}}\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(1+{\mathrm{x}}^2)}^2}{=}_{\mathrm{x}=\tan \theta }\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{1+{\tan }^2\theta }{{(1+{\tan }^2\theta )}^2}\mathrm{d}\theta$$$$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{\cos }^2\theta \mathrm{d}\theta =\frac{1}{4}{\int }_0^{\frac{\pi }{2}}(1+\cos \left(2\theta \right))\mathrm{d}\theta$$$$=\frac{\pi }{8}+\frac{1}{8}{\left[\sin \left(2\theta \right)\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{8}+0=\frac{\pi }{8}$$$$\mathrm{A}=\frac{\pi }{8}$$

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