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Question Number 46369 by Joel578 last updated on 24/Oct/18

$$\int \frac{dx}{x\sqrt{x^2+2x-1}}={\tan }^{-1}\left(\frac{x-1}{\sqrt{x^2+2x-1}}\right)+C$$$$$$$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{confused}\mathrm{in}\mathrm{choosing}\mathrm{the}\mathrm{right}\mathrm{substitution}$$$$\mathrm{so}\mathrm{that}\mathrm{we}\mathrm{can}\mathrm{get}\mathrm{the}\mathrm{result}\mathrm{above}.$$$$\mathrm{Please}\mathrm{help}...$$

Commented by maxmathsup by imad last updated on 24/Oct/18

$$letA=\int \frac{dx}{x\sqrt{x^2+2x-1}}\Rightarrow A=\int \frac{dx}{x\sqrt{x^2+2x+1-2}}=\int \frac{dx}{x\sqrt{{(x+1)}^2-2}}$$$$changementx+1=\sqrt{2}ch\left(t\right)give$$$$A=\int \frac{\sqrt{2}sh\left(t\right)}{(\sqrt{2}ch\left(t\right)-1)\sqrt{2s}h\left(t\right)}dt=\int \frac{dt}{\sqrt{2}\left(\frac{e^t+e^{-t}}{2}\right)-1}$$$$=\int \frac{2dt}{\sqrt{2}(e^t+e^{-t})-2}{=}_{e^t=u}\int \frac{2}{\sqrt{2}(u+u^{-1})-2}\frac{du}{u}$$$$=\int \frac{2du}{\sqrt{2}{u}^2-2u+\sqrt{2}}=\sqrt{2}\int \frac{du}{u^2-\sqrt{2}u+1}$$$$rootsof{u}^2-\sqrt{2}u+1\to {\mathrm{\Delta }}^{{}^{\prime }}=2-1=1\Rightarrow u_1=1+\sqrt{2}and{u}_2=\sqrt{2}-1\Rightarrow$$$$A=\sqrt{2}\int \frac{du}{(u-(1+\sqrt{2})(u-(\sqrt{2}-1))}$$$$=\frac{\sqrt{2}}{2}\int \{\frac{1}{u-(\sqrt{2}+1)}-\frac{1}{u-(\sqrt{2}-1)}\}du$$$$=\frac{\sqrt{2}}{2}ln\mid \frac{u-(\sqrt{2}+1)}{u-(\sqrt{2}-1)}\mid +c\Rightarrow$$$$A=\frac{\sqrt{2}}{2}ln\mid \frac{e^t-\sqrt{2}-1}{u-\sqrt{2}+1}\mid +cbutt=argch\left(\frac{x+1}{\sqrt{2}}\right)=ln(\frac{x+1}{\sqrt{2}}+\sqrt{{\left(\frac{x+1}{\sqrt{2}}\right)}^2-1})\Rightarrow$$$$A=\frac{\sqrt{2}}{2}ln\mid \frac{\frac{x+1}{\sqrt{2}}+\sqrt{{\left(\frac{x+1}{\sqrt{2}}\right)}^2-1}-\sqrt{2}-1}{\frac{x+1}{\sqrt{2}}+\sqrt{{\left(\frac{x+1}{\sqrt{2}}\right)}^2-1}-\sqrt{2}+1}\mid +c.$$$$$$

Answered by Smail last updated on 24/Oct/18

$$I=\int \frac{dx}{x\sqrt{x^2+2x-1}}=\int \frac{dx}{x\sqrt{{(x+1)}^2-2}}$$$$=\int \frac{dx}{x\sqrt{2({\left(\frac{x+1}{\sqrt{2}}\right)}^2-1)}}=\frac{1}{\sqrt{2}}\int \frac{dx}{x\sqrt{{\left(\frac{x+1}{\sqrt{2}}\right)}^2-1}}$$$$t=\frac{x+1}{\sqrt{2}}\Rightarrow dx=\sqrt{2}dt$$$$I=\int \frac{dt}{(\sqrt{2}t-1)\sqrt{t^2-1}}$$$$cosh\left(u\right)=t\Rightarrow dt=sinh\left(u\right)du$$$$I=\int \frac{du}{(\sqrt{2}cosh\left(u\right)-1)}=\int \frac{du}{\sqrt{2}\left(\frac{e^u+e^{-u}}{2}\right)-1}$$$$=2\int \frac{du}{\sqrt{2}(e^u+e^{-u})-2}=2\int \frac{e^{u}du}{\sqrt{2}{e}^{2u}+\sqrt{2}-2e^u}$$$$z=e^u\Rightarrow dz=e^{u}du$$$$I=2\int \frac{dz}{\sqrt{2}{z}^2-2z+\sqrt{2}}=\sqrt{2}\int \frac{dz}{z^2-\sqrt{2}z+1}$$$$=\sqrt{2}\int \frac{dz}{z^2-2\times \frac{\sqrt{2}}{2}z+\frac{1}{2}-\frac{1}{2}+1}$$$$=\sqrt{2}\int \frac{dz}{{(z-\frac{\sqrt{2}}{2})}^2+\frac{1}{2}}=\sqrt{2}\int \frac{dz}{{\left(\frac{\sqrt{2}z-1}{\sqrt{2}}\right)}^2+\frac{1}{2}}$$$$=\sqrt{2}\int \frac{dz}{\frac{1}{2}({\left[\sqrt{2}\left(\frac{\sqrt{2}z-1}{\sqrt{2}}\right)\right]}^2+1)}$$$$=2\sqrt{2}\int \frac{dz}{{(\sqrt{2}z-1)}^2+1}$$$$v=\sqrt{2}z-1\Rightarrow dv=\sqrt{2}dz$$$$I=2\int \frac{dv}{v^2+1}=2{tan}^{-1}\left(v\right)+C$$$$=2{tan}^{-1}(\sqrt{2}z-1)+C=2{tan}^{-1}(\sqrt{2}{e}^u-1)+C$$$$=2{tan}^{-1}(\sqrt{2}{e}^{{cosh}^{-1}\left(t\right)}-1)+C$$$$=2{tan}^{-1}(\sqrt{2}(t+\sqrt{t^2-1})-1)+C$$$$=2{tan}^{-1}(\sqrt{2}(\frac{x+1}{\sqrt{2}}+\sqrt{\frac{{(x+1)}^2}{2}-1})-1)+C$$$$=2{tan}^{-1}(x+1+\sqrt{x^2+2x-1}-1)+C$$$$\int \frac{dx}{x\sqrt{x^2+2x-1}}=2{tan}^{-1}(x+\sqrt{x^2+2x-1})+C$$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Oct/18

$$t=\frac{1}{x}x=\frac{1}{t}dx=\frac{-dt}{t^2}$$$$\int \frac{-dt}{t^2\times \frac{1}{t}\times \sqrt{\frac{1}{t^2}+\frac{2}{t}-1}}$$$$=-1\times \int \frac{dt}{t\sqrt{\frac{1+2t-t^2}{t^2}}}$$$$=-1\times \int \frac{dt}{\sqrt{1-(t^2-2t+1-1)}}$$$$=-1\times \int \frac{dt}{\sqrt{2-{(t-1)}^2}}$$$$=-1\times {sin}^{-1}\left(\frac{t-1}{\sqrt{2}}\right)+c$$$$=-{sim}^{-1}\left(\frac{\frac{1}{x}-1}{\sqrt{2}}\right)+c$$$$=-{tan}^{-1}\left(\frac{t-1}{\sqrt{2-{(t-1)}^2}}\right)$$$$=-{tan}^{-1}\left(\frac{\frac{1}{x}-1}{\sqrt{2-\frac{1}{x^2}+\frac{2}{x}-1}}\right)$$$$=-{tan}^{-1}\left(\frac{1-x}{\sqrt{x^2-1+2x}}\right)$$$$={tan}^{-1}\left(\frac{x-1}{\sqrt{x^2+2x-1}}\right)+c$$

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