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Question Number 133951 by liberty last updated on 25/Feb/21

$$\mathcal{V}=\int \ln (\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})\mathrm{dx}$$

Answered by mathmax by abdo last updated on 25/Feb/21

$$\mathrm{\Phi }=\int \ln (\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=\mathrm{sht}\Rightarrow$$$$\mathrm{\Phi }=\int \ln (\mathrm{sht}+\mathrm{cht})\mathrm{chtdt}=\int \ln (\frac{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}{2}+\frac{{\mathrm{e}}^{\mathrm{t}}+{\mathrm{e}}^{-\mathrm{t}}}{2})\mathrm{ch}\left(\mathrm{t}\right)\mathrm{dt}$$$$=\int \mathrm{tch}\left(\mathrm{t}\right)\mathrm{dt}=\mathrm{tsh}\left(\mathrm{t}\right)-\int \mathrm{sh}\left(\mathrm{t}\right)\mathrm{dt}=\mathrm{tsh}\left(\mathrm{t}\right)-\mathrm{ch}\left(\mathrm{t}\right)+\mathrm{c}$$$$=\mathrm{xargsh}\left(\mathrm{x}\right)-\sqrt{1+{\mathrm{x}}^2}+\mathrm{C}=\mathrm{xln}(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})-\sqrt{1+{\mathrm{x}}^2}+\mathrm{C}$$

Answered by bobhans last updated on 26/Feb/21

$$\mathrm{integration}\mathrm{by}\mathrm{parts}$$$$\{\begin{array}{l}\mathrm{u}=\ln (\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})\to \mathrm{du}=\frac{1+\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}}{\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}}\mathrm{dx}\\ \mathrm{v}=\mathrm{x}\end{array}$$$$\mathrm{I}=\mathrm{xln}(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})-\int \mathrm{x}\left(\frac{\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}}{(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})\sqrt{1+{\mathrm{x}}^2}}\right)\mathrm{dx}$$$$\mathrm{I}=\mathrm{xln}(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})-\int \frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}\mathrm{dx}$$$$\mathrm{I}=\mathrm{xln}(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})-\frac{1}{2}\int \frac{\mathrm{d}(1+{\mathrm{x}}^2)}{\sqrt{1+{\mathrm{x}}^2}}$$$$\mathrm{I}=\mathrm{x}\ln (\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})-\sqrt{1+{\mathrm{x}}^2}+\mathrm{c}$$

Answered by physicstutes last updated on 26/Feb/21

$$\ln (x+\sqrt{1+x^2})={\sinh }^{-}x$$$$\Rightarrow \mathcal{V}=\int {\sinh }^{-1}xdx$$$$\{\begin{array}{l}u={\sinh }^{-1}x\\ dv=dx\end{array}\Rightarrow \{\begin{array}{l}du=\frac{1}{\sqrt{1+x^2}}dx\\ v=x\end{array}$$$$\Rightarrow \mathcal{V}=x{\sinh }^{-1}x-\int \frac{x}{\sqrt{1+x^2}}dx$$$$\mathcal{V}=x{\sinh }^{-1}x-\sqrt{1+x^2}+A,A\in \mathbb{R}$$

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