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Question Number 136948 by leena12345 last updated on 28/Mar/21

$$\int \frac{\sqrt{4+x^2}}{x}dx$$

Answered by Olaf last updated on 28/Mar/21

$$\mathrm{F}\left(x\right)=\int \frac{\sqrt{4+x^2}}{x}dx$$$$\mathrm{F}\left(u\right)\stackrel{x=2\mathrm{sh}u}{=}\int \frac{\sqrt{4+{4\mathrm{sh}}^{2}u}}{2\mathrm{sh}u}2\mathrm{ch}udu$$$$\mathrm{F}\left(u\right)=\int \frac{2\mathrm{ch}u}{2\mathrm{sh}u}2\mathrm{ch}udu$$$$\mathrm{F}\left(u\right)=2\int \frac{{\mathrm{ch}}^{2}u}{\mathrm{sh}u}du$$$$\mathrm{F}\left(u\right)=2\int \frac{{\mathrm{sh}}^{2}u+1}{\mathrm{sh}u}du$$$$\mathrm{F}\left(u\right)=2\int (\mathrm{sh}u+\frac{1}{\mathrm{sh}u})du$$$$\mathrm{F}\left(u\right)=2\int (\mathrm{sh}u-\frac{2e^u}{1-e^{2u}})du$$$$\mathrm{F}\left(u\right)=2[\mathrm{ch}u-2\mathrm{argth}\left(e^u\right)]$$$$\mathrm{F}\left(u\right)=2[\sqrt{{\mathrm{sh}}^{2}u+1}-2\times \frac{1}{2}\ln \mid \frac{1+e^u}{1-e^u}\mid ]$$$$\mathrm{F}\left(u\right)=2[\sqrt{{\mathrm{sh}}^{2}u+1}-2\times \frac{1}{2}\ln \mid \frac{1+\mathrm{shu}+\mathrm{ch}u}{1-\mathrm{sh}u-\mathrm{ch}u}\mid ]$$$$\mathrm{F}\left(x\right)=\sqrt{x^2+4}-2\ln \mid \frac{1+\frac{x}{2}+\sqrt{\frac{x^2}{4}+1}}{1-\frac{x}{2}-\sqrt{\frac{x^2}{4}+1}}\mid (+\mathrm{C})$$$$\mathrm{F}\left(x\right)=\sqrt{x^2+4}-2\ln \mid \frac{2+x+\sqrt{x^2+4}}{2-x-\sqrt{x^2+4}}\mid (+\mathrm{C})$$

Answered by Dwaipayan Shikari last updated on 28/Mar/21

$$x=2sinhu\Rightarrow {xe}^u=e^{2u}-1\Rightarrow e^u=\frac{x\pm \sqrt{x^2+4}}{2}$$$$=\int \frac{4{cosh}^2\left(u\right)}{sinh\left(u\right)}du=4\int \frac{1+{sinh}^{2}u}{sinhu}du$$$$=4\int \frac{1}{e^u-e^{-u}}du+4cosh\left(u\right)=2log\left(\frac{e^u-1}{e^u+1}\right)+4cosh\left(u\right)+C$$$$=2log\left(\frac{x\pm \sqrt{x^2+4}-2}{x\pm \sqrt{x^2+4}+2}\right)+\sqrt{4+x^2}+C$$

Answered by mathmax by abdo last updated on 28/Mar/21

$$\mathrm{\Phi }=\int \frac{\sqrt{4+{\mathrm{x}}^2}}{\mathrm{x}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=2\mathrm{sht}\Rightarrow$$$$\mathrm{\Phi }=\int \frac{2\mathrm{ch}\left(\mathrm{t}\right)}{2\mathrm{sh}\left(\mathrm{t}\right)}\left(2\mathrm{cht}\right)\mathrm{dt}=2\int \frac{{\mathrm{ch}}^2\mathrm{t}}{\mathrm{sht}}\mathrm{dt}$$$$=\int \frac{1+\mathrm{ch}\left(2\mathrm{t}\right)}{\mathrm{sh}\left(\mathrm{t}\right)}\mathrm{dt}=\int \frac{1+\frac{{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{e}}^{-2\mathrm{t}}}{2}}{\frac{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}{2}}\mathrm{dt}=\int \frac{2+{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{e}}^{-2\mathrm{t}}}{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}\mathrm{dt}$$$${=}_{{\mathrm{e}}^{\mathrm{t}}=\mathrm{y}}\int \frac{2+{\mathrm{y}}^2+{\mathrm{y}}^{-2}}{\mathrm{y}-{\mathrm{y}}^{-1}}\frac{\mathrm{dy}}{\mathrm{y}}=\int \frac{2+{\mathrm{y}}^2+{\mathrm{y}}^{-2}}{{\mathrm{y}}^2-1}\mathrm{dy}$$$$=\int \frac{{2\mathrm{y}}^2+{\mathrm{y}}^4+1}{{\mathrm{y}}^4-{\mathrm{y}}^2}\mathrm{dy}=\int \frac{{\mathrm{y}}^4+{2\mathrm{y}}^2+1}{{\mathrm{y}}^2({\mathrm{y}}^2-1)}\mathrm{dy}\mathrm{let}\mathrm{decompose}$$$$\mathrm{F}\left(\mathrm{y}\right)=\frac{{\mathrm{y}}^4+{2\mathrm{y}}^2+1}{{\mathrm{y}}^4-{\mathrm{y}}^2}=\frac{{\mathrm{y}}^4-{\mathrm{y}}^2+{3\mathrm{y}}^2+1}{{\mathrm{y}}^4-{\mathrm{y}}^2}=1+\frac{{3\mathrm{y}}^2+1}{{\mathrm{y}}^4-{\mathrm{y}}^2}$$$$\mathrm{v}\left(\mathrm{y}\right)=\frac{{3\mathrm{y}}^2+1}{{\mathrm{y}}^2({\mathrm{y}}^2-1)}=\frac{{3\mathrm{y}}^2+1}{{\mathrm{y}}^2(\mathrm{y}-1)(\mathrm{y}+1)}=\frac{\mathrm{a}}{\mathrm{y}}+\frac{\mathrm{b}}{{\mathrm{y}}^2}+\frac{\mathrm{c}}{\mathrm{y}-1}+\frac{\mathrm{d}}{\mathrm{y}+1}$$$$\mathrm{b}=-1,\mathrm{c}=\frac{4}{2}=2,\mathrm{d}=\frac{4}{-2}=-2$$$$\mathrm{v}\left(\mathrm{y}\right)=\frac{\mathrm{a}}{\mathrm{y}}-\frac{1}{{\mathrm{y}}^2}+\frac{2}{\mathrm{y}-1}-\frac{2}{\mathrm{y}+1}$$$$\mathrm{v}\left(2\right)=\frac{13}{8}=\frac{\mathrm{a}}{2}-\frac{1}{4}+2-\frac{2}{3}=\frac{\mathrm{a}}{2}-\frac{1}{4}+\frac{4}{3}=\frac{\mathrm{a}}{2}+\frac{13}{12}\Rightarrow \mathrm{a}=...$$$$\int \mathrm{F}\left(\mathrm{y}\right)\mathrm{dy}=\mathrm{y}+\mathrm{aln}\mid \mathrm{y}\mid +\frac{1}{\mathrm{y}}+2\ln \mid \frac{\mathrm{y}-1}{\mathrm{y}+1}\mid +\mathrm{C}$$$$\mathrm{y}={\mathrm{e}}^{\mathrm{t}}\mathrm{ant}\mathrm{t}=\mathrm{argsh}\left(\frac{\mathrm{x}}{2}\right)=\ln (\frac{\mathrm{x}}{2}+\sqrt{1+\frac{{\mathrm{x}}^2}{4}})\Rightarrow$$$$\int \mathrm{F}\left(\mathrm{y}\right)\mathrm{dy}=\frac{\mathrm{x}}{2}+\sqrt{1+\frac{{\mathrm{x}}^2}{4}}+\mathrm{aln}(\frac{\mathrm{x}}{2}+\sqrt{1+\frac{{\mathrm{x}}^2}{2}})$$$$+2\ln \mid \frac{\frac{\mathrm{x}}{2}-1+\sqrt{1+\frac{{\mathrm{x}}^2}{4}}}{\frac{\mathrm{x}}{2}+1+\sqrt{1+\frac{{\mathrm{x}}^2}{4}}}\mid +\mathrm{C}=\mathrm{\Phi }$$

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