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Question Number 163854 by amin96 last updated on 11/Jan/22

$$\mathbf{solution}\mathbf{with}\mathbf{residu}\mathbf{theorem}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathbf{x}}^2}{{\mathbf{x}}^4+2{\mathbf{x}}^2+2}\mathbf{dx}=?$$

Answered by Ar Brandon last updated on 11/Jan/22

$$I={\int }_0^{\mathrm{\infty }}\frac{x^2}{x^4+2x^2+2}dx=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{(x^2+\sqrt{2})+(x^2-\sqrt{2})}{x^4+2x^2+2}dx$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{x^2+\sqrt{2}}{x^4+2x^2+2}dx+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{x^2-\sqrt{2}}{x^4+2x^2+2}dx$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1+\frac{\sqrt{2}}{x^2}}{x^2+2+\frac{2}{x^2}}dx+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1-\frac{\sqrt{2}}{x^2}}{x^2+2+\frac{2}{x^2}}dx$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1+\frac{\sqrt{2}}{x^2}}{{(x-\frac{\sqrt{2}}{x})}^2+2+2\sqrt{2}}dx+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1-\frac{\sqrt{2}}{x^2}}{{(x+\frac{\sqrt{2}}{x})}^2+2-2\sqrt{2}}dx$$$$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{du}{u^2+(2+2\sqrt{2})}+\frac{1}{2}{\int }_{+\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dv}{v^2+2-2\sqrt{2}}$$$$=\frac{1}{2}\cdot \frac{1}{\sqrt{2+2\sqrt{2}}}{\left[\arctan \left(\frac{u}{\sqrt{2+2\sqrt{2}}}\right)\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}=\frac{\pi }{2\sqrt{2+2\sqrt{2}}}$$

Commented by peter frank last updated on 11/Jan/22

$$\mathrm{great}$$

Commented by amin96 last updated on 11/Jan/22

$$greatsir.correctanswer$$

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