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Question Number 115401 by Khalmohmmad last updated on 25/Sep/20

$$\int \frac{\sqrt{\mathrm{x}}}{{\mathrm{x}}^2+1}\mathrm{dx}=?$$

Answered by Dwaipayan Shikari last updated on 25/Sep/20

$$2\int \frac{{\mathrm{t}}^2}{{\mathrm{t}}^4+1}\mathrm{dt}=\int \frac{1}{({\mathrm{t}}^2-\mathrm{i})}+\frac{1}{{\mathrm{t}}^2+\mathrm{i}}=\frac{1}{2\sqrt{\mathrm{i}}}\log \left(\frac{\mathrm{t}-\sqrt{\mathrm{i}}}{\mathrm{t}+\sqrt{\mathrm{i}}}\right)+\frac{1}{\sqrt{\mathrm{i}}}{\tan }^{-1}\frac{\mathrm{t}}{\sqrt{\mathrm{i}}}$$$$$$

Commented by Dwaipayan Shikari last updated on 25/Sep/20

$$\mathrm{Take}\mathrm{limits}$$$${\int }_0^{\mathrm{\infty }}\frac{{2\mathrm{t}}^2}{{\mathrm{t}}^4+1}={\left[\frac{1}{2\sqrt{\mathrm{i}}}\log \left(\frac{\mathrm{t}-\sqrt{\mathrm{i}}}{\mathrm{t}+\sqrt{\mathrm{i}}}\right)\right]}_0^{\mathrm{\infty }}+{\left[\frac{1}{\sqrt{\mathrm{i}}}{\tan }^{-1}\frac{\mathrm{t}}{\sqrt{\mathrm{i}}}\right]}_0^{\mathrm{\infty }}$$$$=\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\frac{1}{2\sqrt{\mathrm{i}}}\log \left(\frac{1-\frac{\sqrt{\mathrm{i}}}{\mathrm{t}}}{1+\frac{\sqrt{\mathrm{i}}}{\mathrm{t}}}\right)-\frac{1}{2\sqrt{\mathrm{i}}}\log (-1)=\frac{1}{2\sqrt{\mathrm{i}}}\log (-1)$$$$=0+\frac{1}{2}\pi \sqrt{\mathrm{i}}(-\log (-1)=\log (-1))$$$$\mathrm{And}$$$${\left[\frac{1}{\sqrt{\mathrm{i}}}{\tan }^{-1}\frac{\mathrm{t}}{\sqrt{\mathrm{i}}}\right]}_0^{\mathrm{\infty }}=\frac{\pi }{2\sqrt{\mathrm{i}}}$$$$\frac{\pi }{2}(\sqrt{\mathrm{i}}+\frac{1}{\sqrt{\mathrm{i}}})=\frac{\pi }{2}\left(\frac{\mathrm{i}+1}{\sqrt{\mathrm{i}}}\right)=\frac{\pi }{2}(\frac{\mathrm{i}+1}{1+\mathrm{i}}.\sqrt{2})=\frac{\pi }{\sqrt{2}}\sqrt{\mathrm{i}}=\pm \frac{1+\mathrm{i}}{\sqrt{2}}$$$$\mathrm{Which}\mathrm{is}\mathrm{Real}$$

Commented by Dwaipayan Shikari last updated on 25/Sep/20

$$\mathrm{It}\mathrm{is}\mathrm{also}\mathrm{same}\mathrm{as}$$$$\int \frac{\sqrt{\mathrm{x}}}{{\mathrm{x}}^2+1}\mathrm{dx}=\int \sqrt{\tan \theta }\mathrm{d}\theta \theta ={\tan }^{-1}\mathrm{x}$$

Answered by TANMAY PANACEA last updated on 25/Sep/20

$$x=k^2$$$$\int \frac{k}{k^4+1}\times 2kdk$$$$\int \frac{2dk}{k^2+\frac{1}{k^2}}$$$$\int \frac{d(k+\frac{1}{k})+d(k-\frac{1}{k})}{k^2+\frac{1}{k^2}}$$$$\int \frac{d(k+\frac{1}{k})}{{(k+\frac{1}{k})}^2-2}+\int \frac{d(k-\frac{1}{k})}{{(k-\frac{1}{k})}^2+2}$$$$nowuseformulapls$$$$\int \frac{dx}{x^2-a^2}and\int \frac{dx}{x^2+a^2}$$

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