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Question Number 74720 by chess1 last updated on 29/Nov/19

Commented by mathmax by abdo last updated on 29/Nov/19

$$changement\sqrt{\frac{x-1}{x+1}}=tgivex-1=t^2(x+1)\Rightarrow (1-t^2)x=1+t^2\Rightarrow$$$$x=\frac{1+t^2}{1-t^2}\Rightarrow \frac{dx}{dt}=\frac{2t(1-t^2)-(1+t^2)(-2t)}{{(1-t^2)}^2}=\frac{2t-2t^3+2t+2t^3}{{(t^2-1)}^2}$$$$=\frac{4t}{{(t^2-1)}^2}\Rightarrow \int \sqrt{\frac{x-1}{x+1}}dx=4\int \frac{t^2}{{(t^2-1)}^2}dtletdecompose$$$$F\left(t\right)=\frac{t^2}{{(t^2-1)}^2}\Rightarrow F\left(t\right)=\frac{t^2}{{(t+1)}^2{(t-1)}^2}$$$$=\frac{a}{t+1}+\frac{b}{{(t+1)}^2}+\frac{c}{t-1}+\frac{d}{{(t-1)}^2}$$$$F(-t)=F\left(t\right)\Rightarrow \frac{-a}{t-1}+\frac{b}{{(t-1)}^2}-\frac{c}{t+1}+\frac{d}{{(t+1)}^2}=\frac{a}{t+1}+\frac{b}{{(t+1)}^2}$$$$+\frac{c}{t-1}+\frac{d}{{(t-1)}^2}\Rightarrow c=-aandb=d\Rightarrow$$$$F\left(t\right)=\frac{a}{t+1}+\frac{b}{{(t+1)}^2}-\frac{a}{t-1}+\frac{b}{{(t-1)}^2}$$$$b={(t+1)}^{2}F\left(t\right){\mid }_{t=-1}=\frac{1}{4}\Rightarrow F\left(t\right)=\frac{a}{t+1}+\frac{1}{4{(t+1)}^2}-\frac{a}{t-1}+\frac{1}{4{(t-1)}^2}$$$$F\left(0\right)=0=a+\frac{1}{4}+a+\frac{1}{4}=2a+\frac{1}{2}\Rightarrow 2a=-\frac{1}{2}\Rightarrow a=-\frac{1}{4}\Rightarrow$$$$F\left(t\right)=\frac{1}{4(t-1)}-\frac{1}{4(t+1)}+\frac{1}{4{(t+1)}^2}+\frac{1}{4{(t-1)}^2}\Rightarrow$$$$\int \sqrt{\frac{x-1}{x+1}}dx=\int \frac{dt}{t-1}-\int \frac{dt}{t+1}+\int \frac{dt}{{(t+1)}^2}+\int \frac{dt}{{(t-1)}^2}$$$$=ln\mid t-1\mid -ln\mid t+1\mid -\frac{1}{t+1}-\frac{1}{t-1}+C$$$$=ln\mid \frac{t-1}{t+1}\mid -(\frac{1}{t+1}+\frac{1}{t-1})+C=ln\mid \frac{t-1}{t+1}\mid -\frac{2t}{t^2-1}+C$$$$=ln\mid \frac{\sqrt{\frac{x-1}{x+1}}-1}{\sqrt{\frac{x-1}{x+1}}+1}\mid -\frac{2\sqrt{\frac{x-1}{x+1}}}{\frac{x-1}{x+1}-1}+C$$$$=ln\mid \frac{\sqrt{\frac{x-1}{x+1}}-1}{\sqrt{\frac{x-1}{x+1}}+1}\mid +(x+1)\sqrt{\frac{x-1}{x+1}}+C.$$$$$$$$$$

Commented by mathmax by abdo last updated on 29/Nov/19

$$wecansimplifyweget$$$$\int \sqrt{\frac{x-1}{x+1}}dx=ln\mid \frac{\sqrt{x-1}-\sqrt{x+1}}{\sqrt{x-1}+\sqrt{x+1}}\mid +\sqrt{x^2-1}+C.$$

Commented by chess1 last updated on 30/Nov/19

$$\mathrm{thanks}$$

Commented by mathmax by abdo last updated on 30/Nov/19

$$youarewelcome.$$

Commented by mathmax by abdo last updated on 06/Dec/19

$$anotherwayweusethechangementx=ch\left(2t\right)\Rightarrow$$$$I=\int \sqrt{\frac{ch\left(2t\right)-1}{ch\left(2t\right)+1}}2sh\left(2t\right)dt=2\int \sqrt{\frac{2{sh}^2\left(t\right)}{2{ch}^2\left(t\right)}}ch\left(2t\right)dt$$$$=2\int th\left(t\right)2sh\left(t\right)ch\left(t\right)dt=4\int \frac{sh\left(t\right)}{ch\left(t\right)}sh\left(t\right)ch\left(t\right)dt$$$$=4\int {sh}^{2}tdt=4\int \frac{ch\left(2t\right)-1}{2}dt=2\int ch\left(2t\right)dt-2t$$$$=sh\left(2t\right)-2t+c=\sqrt{{ch}^2\left(2t\right)-1}-argch\left(x\right)+c$$$$=\sqrt{x^2-1}-ln(x+\sqrt{x^2-1})+c$$

Answered by Tanmay chaudhury last updated on 29/Nov/19

$$\int \frac{x-1}{\sqrt{x^2-1}}dx$$$$\frac{1}{2}\int \frac{d(x^2-1)}{{(x^2-1)}^{\frac{1}{2}}}-\int \frac{dx}{\sqrt{x^2-1}}$$$$I_1=\frac{1}{2}\times \frac{{(x^2-1)}^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+c_1\Rrightarrow {(x^2-1)}^{\frac{1}{2}}+c_1$$$$I_2=\int \frac{dx}{\sqrt{x^2-1}}letx=sec\alpha dx=sec\alpha tan\alpha d\alpha$$$$=\int \frac{sec\alpha tan\alpha d\alpha }{tan\alpha }=\int sec\alpha d\alpha =ln(sec\alpha +tan\alpha )+c_2$$$$=ln(x+\sqrt{x^2-1})+c_2$$

Commented by chess1 last updated on 30/Nov/19

$$\mathrm{thanks}$$

Answered by Kunal12588 last updated on 30/Nov/19

$$I=\int \sqrt{\frac{x-1}{x+1}}dx=\int \frac{x-1}{\sqrt{x^2-1}}dx$$$$=\int \frac{x}{\sqrt{x^2-1}}dx-\int \frac{dx}{\sqrt{x^2-1}}$$$$=\frac{1}{2}\int \frac{2x}{\sqrt{x^2-1}}dx-log\mid x+\sqrt{x^2-1}\mid +C$$$$let{x}^2-1=t\Rightarrow 2xdx=dt$$$$I=\frac{1}{2}\int \frac{dt}{\sqrt{t}}-log\mid x+\sqrt{x^2-1}\mid +C$$$$=\frac{1}{2}\left(\frac{\sqrt{t}}{1/2}\right)-log\mid x+\sqrt{x^2-1}\mid +C$$$$=\sqrt{x^2-1}-log\mid x+\sqrt{x^2-1}\mid +C$$

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