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Question Number 175806 by Ml last updated on 07/Sep/22

Answered by cortano1 last updated on 07/Sep/22

$$\frac{\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}+1}-1}=\frac{\sqrt{\mathrm{x}}(\sqrt{\mathrm{x}+1}+1)}{\mathrm{x}}$$$$=\frac{\sqrt{\mathrm{x}+1}+1}{\sqrt{\mathrm{x}}}$$$$\mathrm{let}\sqrt{\mathrm{x}}=\tan \mathrm{u}\Rightarrow \mathrm{x}=\tan {}^2\left(\mathrm{u}\right)$$$$\{\begin{array}{l}\mathrm{x}=1\Rightarrow \mathrm{u}=\frac{\pi }{4}\\ \mathrm{x}=3\Rightarrow \mathrm{u}=\frac{\pi }{3}\end{array}$$$${\mathrm{I}}_3=\underset{\pi /4}{\stackrel{\pi /3}{\int }}\frac{\sqrt{\tan {}^2\left(\mathrm{u}\right)+1}+1}{\tan \mathrm{u}}\left(2\tan \mathrm{u}\sec {}^2\mathrm{u}\right)\mathrm{du}$$$${\mathrm{I}}_3=\underset{\pi /4}{\stackrel{\pi /3}{\int }}(\sec \mathrm{u}+1)\left(2\sec {}^2\mathrm{u}\right)\mathrm{du}$$$${\mathrm{I}}_3=2\underset{\pi /4}{\stackrel{\pi /3}{\int }}\sec {}^3\mathrm{u}\mathrm{du}+2{\left[\tan \mathrm{u}\right]}_{\pi /4}^{\pi /3}$$$${\mathrm{I}}_3=2(\sqrt{3}-1)+2\underset{\pi /4}{\stackrel{\pi /3}{\int }}\sec \mathrm{u}\mathrm{d}\left(\tan \mathrm{u}\right)$$$$\mathrm{\Omega }=\int \sec \mathrm{u}\mathrm{d}\left(\tan \mathrm{u}\right)=\sec \mathrm{u}\tan \mathrm{u}-\int \sec \mathrm{u}(\sec {}^2\mathrm{u}-1)\mathrm{du}$$$$\mathrm{\Omega }=\sec \mathrm{u}\tan \mathrm{u}-\mathrm{\Omega }+\int \sec \mathrm{u}\mathrm{du}$$$$\mathrm{\Omega }=\frac{1}{2}\sec \mathrm{u}\tan \mathrm{u}+\frac{\ln \mid \sec \mathrm{u}+\tan \mathrm{u}\mid }{2}+\mathrm{c}$$$${\mathrm{I}}_3=2\sqrt{3}-2+2.\frac{1}{2}{[\sec \mathrm{u}\tan \mathrm{u}+\ln \mid \sec \mathrm{u}+\tan \mathrm{u}\mid ]}_{\pi /4}^{\pi /3}$$$${\mathrm{I}}_3=2\sqrt{3}-2+2\sqrt{3}+\ln (2+\sqrt{3})-\sqrt{2}-\ln (1+\sqrt{2})$$$${\mathrm{I}}_3=4\sqrt{3}-2-\sqrt{2}+\ln \left(\frac{2+\sqrt{3}}{\sqrt{2}+1}\right)$$$${\mathrm{I}}_3=4\sqrt{3}-2-\sqrt{2}+\ln \left((2+\sqrt{3})(\sqrt{2}-1)\right)$$$${\mathrm{I}}_3=4\sqrt{3}-2-\sqrt{2}+\ln (2\sqrt{2}-2+\sqrt{6}-\sqrt{3})$$

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