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Question Number 67005 by mathmax by abdo last updated on 21/Aug/19

$$find\int \frac{x-2\sqrt{x^2-1}}{x+2\sqrt{x^2-1}}dx$$

Commented by mathmax by abdo last updated on 27/Aug/19

$$letI=\int \frac{x-2\sqrt{x^2-1}}{x+2\sqrt{x^2-1}}dxchangementx=ch\left(t\right)give$$$$I=\int \frac{ch\left(t\right)-2sh\left(t\right)}{ch\left(t\right)+2sh\left(t\right)}sh\left(t\right)dt=\int \frac{ch\left(t\right)sh\left(t\right)-2{sh}^2\left(t\right)}{ch\left(t\right)+2sh\left(t\right)}dt$$$$=\int \frac{\frac{1}{2}sh\left(2t\right)-2\frac{ch\left(2t\right)-1}{2}}{ch\left(t\right)+2sh\left(t\right)}dt=\frac{1}{2}\int \frac{sh\left(2t\right)-2ch\left(2t\right)+2}{ch\left(t\right)+2sh\left(t\right)}dt$$$$2I=\int \frac{\frac{e^{2t}-e^{-2t}}{2}-2\frac{e^{2t}+e^{-2t}}{2}+2}{\frac{e^t+e^{-t}}{2}+2\frac{e^t-e^{-t}}{2}}dt$$$$=\int \frac{e^{2t}-e^{-2t}-2e^{2t}-2e^{-2t}+4}{e^t+e^{-t}+2e^t-2e^{-t}}dt=\int =\frac{4-e^{2t}-3e^{-2t}}{3e^t-e^{-t}}dt$$$${=}_{e^t=z}\int \frac{4-z^2-3z^{-2}}{3z-z^{-1}}\frac{dz}{z}=\int \frac{4-z^2-3z^{-2}}{3z^2-1}dz$$$$=\int \frac{4z^2-z^4-3}{3z^4-z^2}dzletdecomposeF\left(z\right)=\frac{4z^2-z^4-3}{z^2(3z^2-1)}\Rightarrow$$$$F\left(z\right)=-\frac{z^4-4z^2+3}{3z^4-z^2}=-\frac{1}{3}\frac{3z^4-12z^2+9}{3z^4-z^2}$$$$=-\frac{1}{3}\times \frac{3z^4-z^2-10z^2+9}{3z^4-z^2}=-\frac{1}{3}-\frac{1}{3}\frac{-10z^2+9}{z^2(3z^2-1)}$$$$=-\frac{1}{3}+\frac{a}{z}+\frac{b}{z^2}+\frac{c}{\sqrt{3}z-1}+\frac{d}{\sqrt{3}z+1}\Rightarrow$$$$\int F\left(z\right)dz=-\frac{z}{3}+aln\mid z\mid -\frac{b}{z}+\frac{c}{\sqrt{3}}ln\mid \sqrt{3}z-1\mid +\frac{d}{\sqrt{3}}ln\mid \sqrt{3}z+1\mid +C$$$$=-\frac{e^t}{3}+at-{be}^{-t}+\frac{c}{\sqrt{3}}ln\mid \sqrt{3}{e}^t-1\mid +\frac{d}{\sqrt{3}}ln\mid \sqrt{3}{e}^t+1\mid +C$$$$butt=argch\left(x\right)=ln(x+\sqrt{x^2-1})\Rightarrow$$$$2I=-\frac{1}{3}(x+\sqrt{x^2-1})-\frac{b}{x+\sqrt{x^2-1}}+\frac{c}{\sqrt{3}}ln\mid \sqrt{3}(x+\sqrt{x^2-1})-1\mid$$$$+\frac{d}{\sqrt{3}}ln\mid \sqrt{3}(x+\sqrt{x^2-1})+1\mid +C$$$$resttocalculatethecoeff.c_i.....$$

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