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Question Number 66309 by mathmax by abdo last updated on 12/Aug/19

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

Commented by prof Abdo imad last updated on 15/Aug/19

$$letI=\int \frac{dx}{(x^2-1)\sqrt{x^2+2}}changement$$$$x=\sqrt{2}sh\left(t\right)giveI=\int \frac{\sqrt{2}ch\left(t\right)}{(2{sh}^{2}t-1)\sqrt{2}ch\left(t\right)}dt$$$$=\int \frac{dt}{2\frac{ch\left(2t\right)-1}{2}-1}=\int \frac{dt}{ch\left(2t\right)-2}$$$$=\int \frac{dt}{\frac{e^{2t}+e^{-2t}}{2}-2}=\int \frac{2dt}{e^{2t}+e^{-2t}-4}$$$${=}_{e^{2t}=u}\int \frac{2}{u+u^{-1}-4}\frac{du}{2u}=\int \frac{du}{u^2+1-4u}$$$$=\int \frac{du}{u^2-4u+1}$$$$u^2-4u+1=0\to {\mathrm{\Delta }}^{{}^{\prime }}=4-1=3\Rightarrow u_1=2+\sqrt{3}$$$$u_2=2-\sqrt{3}\Rightarrow I=\frac{1}{2\sqrt{3}}\int (\frac{1}{u-u_1}-\frac{1}{u-u_2})du$$$$=\frac{1}{2\sqrt{3}}ln\mid \frac{u-u_1}{u-u_2}\mid +c=\frac{1}{2\sqrt{3}}ln\mid \frac{u-2-\sqrt{3}}{u-2+\sqrt{3}}\mid +c$$$$\frac{1}{2\sqrt{3}}ln\mid \frac{e^{2t}-2-\sqrt{3}}{e^{2t}-2+\sqrt{3}}\mid +cbutwehave$$$$t=argsh\left(\frac{x}{\sqrt{2}}\right)=ln(\frac{x}{\sqrt{2}}+\sqrt{1+\frac{x^2}{2}})\Rightarrow$$$$e^{2t}={(\frac{x}{\sqrt{2}}+\sqrt{1+\frac{x^2}{2}})}^2\Rightarrow$$$$I=\frac{1}{2\sqrt{3}}ln\mid \frac{\frac{x+\sqrt{x^2+2}}{\sqrt{2}}-2-\sqrt{3}}{\frac{x+\sqrt{x^2+2}}{\sqrt{2}}-2+\sqrt{3}}\mid +c$$$$$$$$$$$$$$

Commented by prof Abdo imad last updated on 15/Aug/19

$$I=\frac{1}{2\sqrt{3}}ln\mid \frac{{\left(\frac{x+\sqrt{x^2+2}}{\sqrt{2}}\right)}^2-2-\sqrt{3}}{{\left(\frac{x+\sqrt{x^2+2}}{\sqrt{2}}\right)}^2-2+\sqrt{3}}\mid +c$$

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