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Question Number 39020 by maxmathsup by imad last updated on 01/Jul/18

$$calculate{\int }_0^1\frac{ln(1+\sqrt{x^2+1})}{\sqrt{x^2+1}}dx$$

Answered by behi83417@gmail.com last updated on 02/Jul/18

$$x=tgt\Rightarrow dx=(1+{tg}^{2}t)dt$$$$I=\int \frac{ln(1+sect)}{sect}.{sec}^{2}tdt=$$$$=\int sect.ln(1+sect)dt=$$$$sect.ln(1+sect)-\int sect.\frac{sint.{sec}^{2}t}{1+sect}dt=$$$$=do-\int \frac{sint.\frac{1}{{cos}^{3}t}}{1+\frac{1}{cost}}dt=do-\int \frac{sintdt}{{cos}^{3}t+{cos}^{2}t}$$$$cost=u\Rightarrow \int \frac{-sintdt}{{cos}^{3}t+{cos}^{2}t}=\int \frac{du}{u^3+u^2}=$$$$=\int \frac{du}{u^2(u+1)}=\int [\frac{-1}{u}+\frac{1}{u^2}+\frac{1}{u+1}]du$$$$=-lnu-u+ln(u+1)+c=$$$$=-lncost-cost+ln(1+cost)+c=$$$$=ln\frac{1+cost}{cost}-cost+c=ln(1+sect)-cost+c$$$$\Rightarrow I=\sqrt{1+{\mathit{x}}^2}.\mathit{l}\mathit{n}(1+\sqrt{1+{\mathit{x}}^2})+\mathit{l}\mathit{n}(1+\sqrt{1+{\mathit{x}}^2})-\frac{1}{\sqrt{1+{\mathit{x}}^2}}+\mathit{c}=$$$$=(1+\sqrt{1+{\mathit{x}}^2})\mathit{l}\mathit{n}(1+\sqrt{1+{\mathit{x}}^2})-\frac{1}{\sqrt{1+{\mathit{x}}^2}}+\mathit{c}.◼$$

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