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Question Number 145183 by mathmax by abdo last updated on 03/Jul/21

$$\mathrm{find}{\int }_0^1\frac{\mathrm{dx}}{{(\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}+1})}^3}$$

Commented by justtry last updated on 03/Jul/21

Answered by mindispower last updated on 03/Jul/21

$$x={sh}^2\left(t\right)$$$$\iff \int \frac{sh\left(2t\right)dt}{e^{3t}}$$$$=\frac{1}{2e^t}-\frac{e^{-5t}}{2}$$$$=-\frac{e^{-t}}{2}+\frac{e^{-5t}}{10}+c,t=argsh\left(\sqrt{x}\right)$$

Answered by mathmax by abdo last updated on 03/Jul/21

$$\mathrm{\Psi }={\int }_0^1\frac{\mathrm{dx}}{{(\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}+1})}^3}\mathrm{changement}\mathrm{x}={\mathrm{sh}}^2\mathrm{t}\mathrm{give}$$$$\mathrm{sht}=\sqrt{\mathrm{x}}\Rightarrow \mathrm{t}=\mathrm{argsh}\left(\sqrt{\mathrm{x}}\right)=\log (\sqrt{\mathrm{x}}+\sqrt{1+\mathrm{x}})\Rightarrow$$$$\mathrm{\Psi }={\int }_0^{\log (1+\sqrt{2})}\frac{2\mathrm{sht}\mathrm{cht}}{{(\mathrm{sht}+\mathrm{cht})}^3}\mathrm{dt}={\int }_0^{\log (1+\sqrt{2})}\frac{\mathrm{sh}\left(2\mathrm{t}\right)}{{\mathrm{e}}^{3\mathrm{t}}}\mathrm{dt}$$$$={\int }_0^{\log (1+\sqrt{2})}{\mathrm{e}}^{-3\mathrm{t}}.\frac{{\mathrm{e}}^{2\mathrm{t}}-{\mathrm{e}}^{-2\mathrm{t}}}{2}\mathrm{dt}=\frac{1}{2}{\int }_0^{\log (1+\sqrt{2})}({\mathrm{e}}^{-\mathrm{t}}-{\mathrm{e}}^{-5\mathrm{t}})\mathrm{dt}$$$$=\frac{1}{2}{[-{\mathrm{e}}^{-\mathrm{t}}+\frac{1}{5}{\mathrm{e}}^{-5\mathrm{t}}]}_0^{\log (1+\sqrt{2})}$$$$\mathrm{\Psi }=\frac{1}{2}(-\frac{1}{1+\sqrt{2}}+\frac{1}{5{(1+\sqrt{2})}^5}+1-\frac{1}{5})$$

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