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Question Number 200417 by hardmath last updated on 18/Nov/23

$$\mathrm{find}:\mathrm{\Omega }={\int }_1^{\mathrm{\infty }}\frac{\sqrt{\mathrm{x}}}{(1+{\mathrm{x}}^2)}\mathrm{dx}=?$$

Answered by witcher3 last updated on 18/Nov/23

$$\sqrt{\mathrm{x}}=\mathrm{y}$$$$={\int }_1^{\mathrm{\infty }}\frac{{2\mathrm{y}}^2}{(1+{\mathrm{y}}^4)}\mathrm{dy}$$$$={\int }_1^{\mathrm{\infty }}\frac{{2\mathrm{y}}^2}{({\mathrm{y}}^2+\mathrm{y}\sqrt{2}+1)({\mathrm{y}}^2-\mathrm{y}\sqrt{2}+1)}$$$$=\frac{1}{\sqrt{2}}{\int }_1^{\mathrm{\infty }}-\frac{\mathrm{y}}{{\mathrm{y}}^2+\mathrm{y}\sqrt{2}+1}+\frac{\mathrm{y}}{{\mathrm{y}}^2-\mathrm{y}\sqrt{2}+1}\mathrm{dy}$$$$$$$$=\frac{1}{2\sqrt{2}}{\int }_1^{\mathrm{\infty }}\frac{2\mathrm{y}-\sqrt{2}+\sqrt{2}}{{\mathrm{y}}^2-\mathrm{y}\sqrt{2}+1}-\frac{2\mathrm{y}+\sqrt{2}-\sqrt{2}}{{\mathrm{y}}^2+\mathrm{y}\sqrt{2}+1}\mathrm{dy}$$$$=\frac{1}{2\sqrt{2}}\ln \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)+\frac{1}{2}{\int }_1^{\mathrm{\infty }}\frac{\mathrm{dy}}{{(\mathrm{y}-\frac{1}{\sqrt{2}})}^2+\frac{1}{2}}+\frac{1}{2}{\int }_1^{\mathrm{\infty }}\frac{\mathrm{dy}}{{(\mathrm{y}+\frac{1}{\sqrt{2}})}^2+\frac{1}{2}}$$$$=\frac{\ln \left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right)}{2\sqrt{2}}+\frac{1}{\sqrt{2}}{[{\tan }^{-1}(\mathrm{y}\sqrt{2}-1)+{\tan }^{-1}(\mathrm{y}\sqrt{2}+1)]}_1^{\mathrm{\infty }}$$$$=\frac{\ln \left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right)}{2\sqrt{2}}+\frac{1}{\sqrt{2}}[{\tan }^{-1}(\sqrt{2}-1)+{\tan }^{-1}(\sqrt{2}+1)]$$$${\tan }^{-1}(\sqrt{2}+1)+{\tan }^{-1}(\sqrt{2}-1)=\frac{\pi }{2}$$$$\mathrm{we}\mathrm{get}$$$$\frac{\pi }{2\sqrt{2}}+\frac{\ln (1+\sqrt{2)}}{\sqrt{2}}$$$$$$

Commented by hardmath last updated on 19/Nov/23

$$thankyoudearprofessor$$

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