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Question Number 162811 by mnjuly1970 last updated on 01/Jan/22

$$$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{tan}^{-1}\left(x\right)}{{(1+x^2)}^2}dx=?$$$$----------$$

Answered by amin96 last updated on 01/Jan/22

$$$$$$\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{g}\left(\mathit{x}\right)=\mathit{t}\mathbf{x}=\mathbf{tg}\left(\mathbf{t}\right)\frac{\mathit{d}\mathit{t}}{\mathit{d}\mathit{x}}=\frac{1}{{\mathit{x}}^2+1}t[\frac{\pi }{2}:0]$$$$I={\int }_0^{\frac{\pi }{2}}\frac{\mathit{t}}{(1+{\mathit{x}}^2)}\mathit{d}\mathit{t}={\int }_0^{\frac{\pi }{2}}\frac{\mathbf{t}}{1+{\mathbf{tg}}^2\left(\mathbf{t}\right)}\mathbf{dt}={\int }_0^{\frac{\pi }{2}}{\mathit{t}\mathit{c}\mathit{o}\mathit{s}}^2\left(\mathit{t}\right)\mathit{d}\mathit{t}$$$$\mathbf{IBP}\mathit{t}=\mathit{u}\mathit{d}\mathit{u}=\mathit{d}\mathit{t}\mathit{v}=\frac{2\mathit{t}+\mathit{s}\mathit{i}\mathit{n}\left(2\mathit{t}\right)}{4}$$$$\mathit{I}={\left[\mathit{t}(\frac{\mathit{t}}{2}+\frac{\mathit{s}\mathit{i}\mathit{n}\left(2\mathit{t}\right)}{4})\right]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}(\frac{\mathit{t}}{2}+\frac{\mathit{s}\mathit{i}\mathit{n}\left(2\mathit{t}\right)}{4})\mathit{d}\mathit{t}=$$$$=\frac{{\pi }^2}{8}-{\left[\frac{{\mathit{t}}^2}{4}\right]}_0^{\frac{\pi }{2}}-\frac{1}{4}\underset{1}{\underbrace{{\int }_0^{\frac{\pi }{2}}\mathit{s}\mathit{i}\mathit{n}\left(2\mathit{t}\right)\mathit{d}\mathit{t}}}=\frac{{\pi }^2}{8}-\frac{{\pi }^2}{16}-\frac{1}{4}=$$$$=\frac{{\pi }^2}{16}-\frac{1}{4}$$$$$$

Commented by mnjuly1970 last updated on 01/Jan/22

$$yessir?amin.thankyousomuch$$

Commented by amin96 last updated on 01/Jan/22

$$$$Thanks for the good questions too. I'm your follower👍 sir MN

Commented by mnjuly1970 last updated on 01/Jan/22

$$that,smypleasuresiramin.$$$$thankyouverymuch..$$

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