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Question Number 143100 by lapache last updated on 10/Jun/21

$$\int \sqrt{e^x+1}=.....???$$

Answered by qaz last updated on 10/Jun/21

$$\int \sqrt{{\mathrm{e}}^{\mathrm{x}}+1}\mathrm{dx}$$$$=\int \sqrt{\mathrm{u}+1}\frac{\mathrm{du}}{\mathrm{u}}$$$$=\int \frac{\mathrm{u}+1}{\mathrm{u}\sqrt{\mathrm{u}+1}}\mathrm{du}$$$$=2\sqrt{\mathrm{u}+1}+\int \frac{\mathrm{du}}{{\mathrm{u}}^2\sqrt{\frac{1}{\mathrm{u}}+\frac{1}{{\mathrm{u}}^2}}}$$$$=2\sqrt{\mathrm{u}+1}-\int \frac{\mathrm{d}\left(\frac{1}{\mathrm{u}}\right)}{\sqrt{{(\frac{1}{\mathrm{u}}+\frac{1}{2})}^2-\frac{1}{4}}}$$$$=2\sqrt{\mathrm{u}+1}-\int \frac{\mathrm{d}\left[2(\frac{1}{\mathrm{u}}+\frac{1}{2})\right]}{\sqrt{[4{(\frac{1}{\mathrm{u}}+\frac{1}{2})}^2-1]}}$$$$=2\sqrt{\mathrm{u}+1}-\mathrm{arccosh}\left(2(\frac{1}{\mathrm{u}}+\frac{1}{2})\right)+\mathrm{C}$$$$=2\sqrt{{\mathrm{e}}^{\mathrm{x}}+1}-\mathrm{arccosh}\left(\frac{2+{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}}\right)+\stackrel{}{\mathrm{C}}$$

Answered by MJS_new last updated on 10/Jun/21

$$\int \sqrt{{\mathrm{e}}^x+1}dx=$$$$[t=\sqrt{{\mathrm{e}}^x+1}\to dx=\frac{2\sqrt{{\mathrm{e}}^x+1}}{{\mathrm{e}}^x}dt]$$$$=2\int \frac{t^2}{t^2-1}dt=\int (2+\frac{1}{t-1}-\frac{1}{t+1})dt=$$$$=2t+\ln \frac{t-1}{t+1}=...$$$$=2\sqrt{{\mathrm{e}}^x+1}-x+\ln ({\mathrm{e}}^x+2-2\sqrt{{\mathrm{e}}^x+1})+C$$

Answered by pticantor last updated on 10/Jun/21

$$wokoooettuveuxlerizpourcassersahein$$$$tuvaslire?l^{\prime }heure$$

Answered by mathmax by abdo last updated on 11/Jun/21

$$\mathrm{I}=\int \sqrt{1+{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\sqrt{1+{\mathrm{e}}^{\mathrm{x}}}=\mathrm{t}\Rightarrow$$$$1+{\mathrm{e}}^{\mathrm{x}}={\mathrm{t}}^2\Rightarrow {\mathrm{e}}^{\mathrm{x}}={\mathrm{t}}^2-1\Rightarrow \mathrm{x}=\log ({\mathrm{t}}^2-1)\Rightarrow$$$$\mathrm{I}=\int \mathrm{t}\times \frac{2\mathrm{t}}{{\mathrm{t}}^2-1}\mathrm{dt}=2\int \frac{{\mathrm{t}}^2}{{\mathrm{t}}^2-1}\mathrm{dt}=2\int \frac{{\mathrm{t}}^2-1+1}{{\mathrm{t}}^2-1}\mathrm{dt}$$$$=2\mathrm{t}+\int \frac{2\mathrm{dt}}{{\mathrm{t}}^2-1}=2\mathrm{t}+\int (\frac{1}{\mathrm{t}-1}-\frac{1}{\mathrm{t}+1})\mathrm{dt}=2\mathrm{t}+\log \mid \frac{\mathrm{t}-1}{\mathrm{t}+1}\mid +\mathrm{C}$$$$\mathrm{I}=2\sqrt{1+{\mathrm{e}}^{\mathrm{x}}}+\log \mid \frac{\sqrt{1+{\mathrm{e}}^{\mathrm{x}}}-1}{\sqrt{1+{\mathrm{e}}^{\mathrm{x}}+1}}\mid +\mathrm{C}$$

Commented by mathmax by abdo last updated on 11/Jun/21

$$\mathrm{typo}\mathrm{I}=2\sqrt{1+{\mathrm{e}}^{\mathrm{x}}}+\log \mid \frac{\sqrt{1+{\mathrm{e}}^{\mathrm{x}}}-1}{\sqrt{1+{\mathrm{e}}^{\mathrm{x}}}+1}\mid +\mathrm{C}$$

Answered by puissant last updated on 23/Aug/21

$$u=\sqrt{e^x+1}\to e^x=u^2-1\to x=ln(u^2-1)$$$$dx=\frac{2u}{u^2-1}du$$$$K=2\int \frac{u^2}{u^2-1}du$$$$K=2\int du+2\int \frac{1}{(u-1)(u+1)}du$$$$\Rightarrow K=2u+\int \frac{1}{u-1}du-\int \frac{1}{u+1}du$$$$\Rightarrow K=2u+ln\mid u-1\mid -ln\mid u+1\mid +C$$$$$$$$\therefore \because K=2\sqrt{e^x+1}+ln\mid \frac{\sqrt{e^x+1}-1}{\sqrt{e^x+1}+1}\mid +C..$$

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