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Question Number 81888 by rajesh4661kumar@gmail.com last updated on 16/Feb/20

Commented by mathmax by abdo last updated on 16/Feb/20

$$letA=\int \frac{dx}{(x^2+1)\sqrt{x^2-1}}wedothechangementx=cht\Rightarrow$$$$A=\int \frac{shtdt}{({ch}^{2}t+1)sht}=\int \frac{dt}{1+\frac{1+ch\left(2t\right)}{2}}=\int \frac{2dt}{3+ch\left(2t\right)}$$$${=}_{2t=u}\int \frac{du}{3+ch\left(u\right)}=\int \frac{du}{3+\frac{e^u+e^{-u}}{2}}=\int \frac{2du}{6+e^u+e^{-u}}$$$${=}_{e^u=z}\int \frac{2}{6+z+z^{-1}}\frac{dz}{z}=\int \frac{2dz}{6z+z^2+1}=\int \frac{2dz}{z^2+6z+1}$$$$z^2+6z+1=0\to {\mathrm{\Delta }}^{{}^{\prime }}=9-1=8\Rightarrow z_1=-3+2\sqrt{2}and{z}_2=-3-2\sqrt{2}\Rightarrow$$$$A=\int \frac{2dz}{(z-z_1)(z-z_2)}=\frac{2}{z_1-z_2}\int (\frac{1}{z-z_1}-\frac{1}{z-z_2})dz$$$$=\frac{2}{4\sqrt{2}}ln\mid \frac{z-z_1}{z-z_2}\mid +C=\frac{1}{2\sqrt{2}}ln\mid \frac{z+3-2\sqrt{2}}{z+3+2\sqrt{2}}\mid +C$$$$=\frac{1}{2\sqrt{2}}ln\mid \frac{e^u+3-2\sqrt{2}}{e^u+3+2\sqrt{2}}\mid +C=\frac{1}{2\sqrt{2}}ln\mid \frac{e^{2t}+3-2\sqrt{2}}{e^{2t}+3+2\sqrt{2}}\mid +C$$$$wehavet=argch\left(x\right)=ln(x+\sqrt{x^2-1})\Rightarrow 2t=ln\left({(x+\sqrt{x^2-1})}^2\right)$$$$\Rightarrow e^{2t}={(x+\sqrt{x^2-1})}^2=x^2+2x\sqrt{x^2-1}+x^2-1$$$$=2x^2-1+2x\sqrt{x^2-1}\Rightarrow$$$$A=\frac{1}{2\sqrt{2}}ln\mid \frac{2x^2+2x\sqrt{x^2-1}+2-2\sqrt{2}}{2x^2+2x\sqrt{x^2-1}+2+2\sqrt{3}}\mid +C$$$$=\frac{1}{2\sqrt{2}}ln\mid \frac{x^2+x\sqrt{x^2-1}+1-\sqrt{2}}{x^2+x\sqrt{x^2-1}+1+\sqrt{2}}\mid +C$$

Answered by TANMAY PANACEA last updated on 16/Feb/20

$$t^2=\frac{x^2-1}{x^2+1}\to x^2{t}^2+t^2=x^2-1$$$$x^2(t^2-1)=-t^2-1$$$$x^2=\frac{1+t^2}{1-t^2}\to x=\sqrt{\frac{1+t^2}{1-t^2}}$$$$dx=\frac{1}{2}{\left(\frac{1+t^2}{1-t^2}\right)}^{\frac{-1}{2}}\times \frac{(1-t^2).2t-(1+t^2)(-2t)}{{(1-t^2)}^2}dt$$$$dx=\frac{1}{2}\times \sqrt{\frac{1-t^2}{1+t^2}}\times \frac{2t-2t^3+2t+2t^3}{{(1-t^2)}^2}dt$$$$\int \frac{2tdt}{\sqrt{1+t^2}\times {(1-t^2)}^{\frac{3}{2}}}\times \frac{1}{(\frac{1+t^2}{1-t^2}+1)(\sqrt{\frac{1+t^2}{1-t^2}-1}}$$$$\int \frac{2tdt}{\sqrt{1+t^2}{(1-t^2)}^{\frac{3}{2}}\times \frac{2}{1-t^2}\times \sqrt{\frac{1+t^2-1+t^2}{1-t^2}}}$$$$\int \frac{2tdt}{\sqrt{1+t^2}}\times \frac{1}{\sqrt{2}t}$$$$\sqrt{2}\int \frac{dt}{\sqrt{1+t^2}}=\sqrt{2}ln(t+\sqrt{1+t^2})+c$$$$now\sqrt{2}ln(\sqrt{\frac{x^2-1}{x^2+1}}+\sqrt{1+\frac{x^2-1}{x^2+1}})+c$$$$\sqrt{2}ln\left(\frac{\sqrt{x^2-1}+x\sqrt{2}}{\sqrt{x^2+1}}\right)+c$$$$\mathit{p}\mathit{l}\mathit{s}\mathit{c}\mathit{h}\mathit{e}\mathit{c}\mathit{k}...$$$$$$

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