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Question Number 173149 by mnjuly1970 last updated on 07/Jul/22

$$$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{\sqrt{1+x}.(2+2x+x^2)}=\frac{1}{\sigma }(\pi -\ln (3+2\sqrt{3}))$$$$\sigma =?$$$$--\mathrm{solution}--$$$$\mathrm{\Omega }\stackrel{\sqrt{1+x}=t}{=}2{\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^4}=2{\int }_0^1\frac{dt}{1+t^4}$$$$\mathrm{\Psi }={\int }_0^1\frac{dt}{1+t^4}=\frac{1}{2}{\int }_0^1\frac{1+t^2-(t^{-2}-1)}{1+t^4}dt$$$$=\frac{1}{2}{\int }_0^1\frac{1+t^2}{1+t^4}dt+\frac{1}{2}{\int }_0^1\frac{1-t^2}{1+t^4}dt$$$$\mathrm{\Phi }={\int }_0^1\frac{1+t^2}{1+t^4}dt={\int }_0^1\frac{t^{-2}+1}{t^{-2}+t^2}dt$$$$={\int }_0^1\frac{1+t^{-2}}{{(t^{}-t^{-1})}^2+2}\stackrel{sub}{=}{\left[\frac{1}{\sqrt{2}}{tan}^{-1}(t-t^{-1})\right]}_0^1=\frac{\pi }{2\sqrt{2}}\ast$$$$\mathit{\varphi }={\int }_0^1\frac{1-t^2}{1+t^4}dt={\int }_0^1\frac{t^{-2}-1}{{(t+t^{-1})}^2-2}dt$$$$\stackrel{t+\frac{1}{t}=u}{=}-{\int }_2^{\mathrm{\infty }}\frac{du}{(u-\sqrt{2})(u+\sqrt{2})}$$$$=\frac{-1}{2\sqrt{2}}{\left[\ln \left(\frac{u-\sqrt{2}}{u+\sqrt{2}}\right)\right]}_2^{\mathrm{\infty }}=\frac{1}{2\sqrt{2}}\ln \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)$$$$\mathit{\varphi }=-\frac{1}{2\sqrt{2}}\ln (3+2\sqrt{2})\ast \ast$$$$(\ast )\mathrm{\&}(\ast \ast )::\mathrm{\Omega }=2\mathrm{\Psi }=(\mathrm{\Phi }+\mathit{\varphi })=\frac{1}{2\sqrt{2}}(\pi -\ln (3+2\sqrt{2})...◼\mathrm{m}.\mathrm{n}$$$$$$

Commented by Tawa11 last updated on 11/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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