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Question Number 85596 by M±th+et£s last updated on 23/Mar/20

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

Commented by mathmax by abdo last updated on 23/Mar/20

$$A=\int \frac{\sqrt{x+1}-1}{\sqrt{x-1}+1}dxchagement\sqrt{x-1}+1=tgive\sqrt{x-1}=t-1\Rightarrow$$$$x-1={(t-1)}^2\Rightarrow dx=2(t-1)dt\Rightarrow$$$$A=\int \frac{\sqrt{1+{(t-1)}^2}-1}{t}2(t-1)dt=2\int \frac{t-1}{t}\sqrt{{(t-1)}^2+1}dt$$$$=2\int (1-\frac{1}{t})\sqrt{{(t-1)}^2+1}dt=2\int \sqrt{{(t-1)}^2+1}dt-2\int \frac{\sqrt{{(t-1)}^2+1}}{t}dt$$$$\int \sqrt{1+{(t-1)}^2}dt{=}_{t-1=sh\left(u\right)}\int {ch}^2\left(u\right)du$$$$=\frac{1}{2}\int (1+ch\left(2u\right))du=\frac{u}{2}+\frac{1}{4}sh\left(2u\right)+c_1$$$$=\frac{u}{2}+\frac{1}{2}sh\left(u\right)ch\left(u\right)+c_1=\frac{1}{2}ln(t-1+\sqrt{1+{(t-1)}^2})$$$$+\frac{1}{2}(t-1)\sqrt{1+{(t-1)}^2}+c_1=\frac{1}{2}ln(\sqrt{x-1}+\sqrt{1+x-1})$$$$+\frac{1}{2}\sqrt{x-1}\sqrt{1+x-1}+c_1=\frac{1}{2}ln(\sqrt{x-1}+\sqrt{x})+\frac{1}{2}\sqrt{x-1}\sqrt{x}+c_1$$$$\int \frac{\sqrt{1+{(t-1)}^2}}{t}dt{=}_{t-1=sh\left(u\right)}\int \frac{chu}{1+sh\left(u\right)}ch\left(u\right)du$$$$=\frac{1}{2}\int \frac{(1+ch\left(2u\right))}{1+sh\left(u\right)}du=\frac{1}{2}\int \frac{(1+\frac{e^{2u}+e^{-2u}}{2})}{1+\frac{e^u-e^{-u}}{2}}du$$$$=\frac{1}{2}\int \frac{2+e^{2u}+e^{-2u}}{2+e^u-e^{-u}}du=\frac{1}{2}\int \frac{2e^{2u}+e^{4u}+1}{2e^{2u}+e^{3u}-e^u}du$$$${=}_{e^u=z}\frac{1}{2}\int \frac{z^4+2z^2+1}{z^3+2z^2-z}\frac{dz}{z}=\frac{1}{2}\int \frac{z^4+2z^2+1}{z^4+2z^3-z^2}dz$$$$=\frac{1}{2}\int \frac{z^4+2z^3-z^2+2z^2+1-2z^3+z^2}{z^4+2z^3-z^2}dz$$$$=\frac{1}{2}\int dz+\frac{1}{2}\int \frac{-2z^3+3z^2+1}{z^4+2z^3-z^2}dzletdecompose$$$$F\left(z\right)=\frac{-2z^3+3z^2+1}{z^2(z^2+2z-1)}\Rightarrow F\left(z\right)=\frac{-2z^3+3z^2+1}{z^2({(z+1)}^2-2)}$$$$=\frac{-2z^3+3z^2+1}{z^2(z+1-\sqrt{2})(z+1+\sqrt{2})}=\frac{a}{z}+\frac{b}{z^2}+\frac{c}{z+1-\sqrt{2}}+\frac{d}{z+1+\sqrt{2}}$$$$\Rightarrow \int F\left(z\right)dz=aln\mid z\mid -\frac{b}{z}+cln\mid z+1-\sqrt{2}\mid +dln\mid z+1+\sqrt{2}\mid +C$$$$restcalculusofcoefgicients...becontinued..$$

Answered by MJS last updated on 23/Mar/20

$$\int \frac{\sqrt{x+1}-1}{\sqrt{x-1}+1}dx=$$$$[t=\sqrt{x-1}\to dx=2\sqrt{x-1}dt]$$$$=2\int \frac{t\sqrt{t^2+2}-t}{t+1}dt=$$$$=-2\int \frac{\sqrt{t^2+2}}{t+1}dt+2\int \sqrt{t^2+2}dt+2\int \frac{dt}{t+1}-2\int dt$$$$$$$$-2\int dt=-2t=-2\sqrt{x-1}\bullet$$$$2\int \frac{dt}{t+1}=2\ln (t+1)=2\ln (\sqrt{x-1}+1)\bullet$$$$2\int \sqrt{t^2+2}dt=t\sqrt{t^2+2}+2\ln (t+\sqrt{t^2+2})=$$$$=\sqrt{x-1}\sqrt{x+1}+2\ln (\sqrt{x-1}+\sqrt{x+1})\bullet$$$$$$$$-2\int \frac{\sqrt{t^2+2}}{t+1}dt=$$$$\mathrm{I}\mathrm{prefer}t=\sqrt{2}\sinh \left(\ln u\right)\mathrm{to}\mathrm{avoid}\mathrm{another}\mathrm{step}$$$$[u=\frac{t+\sqrt{t^2+2}}{\sqrt{2}}\to dt=\frac{\sqrt{2(t^2+2)}}{t+\sqrt{t^2}+2}du=\frac{u^2+1}{\sqrt{2}{u}^2}du]$$$$=-\sqrt{2}\int \frac{{(u^2+1)}^2}{u^2(u^2+\sqrt{2}u-1)}du=$$$$=-\sqrt{2}\int \frac{{(u^2+1)}^2}{u^2(u+\frac{\sqrt{2}+\sqrt{6}}{2})(u+\frac{\sqrt{2}-\sqrt{6}}{2})}du=$$$$=2\sqrt{3}\int \frac{du}{u+\frac{\sqrt{2}+\sqrt{6}}{2}}-2\sqrt{3}\int \frac{du}{u+\frac{\sqrt{2}-\sqrt{6}}{2}}+2\int \frac{du}{u}+\sqrt{2}\int \frac{du}{u^2}-\sqrt{2}\int du=$$$$=2\sqrt{3}\ln (u+\frac{\sqrt{2}+\sqrt{6}}{2})-2\sqrt{3}\ln (u+\frac{\sqrt{2}-\sqrt{6}}{2})+2\ln u-\frac{\sqrt{2}}{u}-\sqrt{2}u=$$$$=2\sqrt{3}\ln \frac{2u+\sqrt{2}+\sqrt{6}}{2u+\sqrt{2}-\sqrt{6}}+2\ln u-\frac{\sqrt{2}(u^2+1)}{u}=$$$$=2\sqrt{3}\ln \frac{2(\sqrt{x-1}+\sqrt{x+1})+\sqrt{2}+\sqrt{6}}{2(\sqrt{x-1}+\sqrt{x+1})+\sqrt{2}-\sqrt{6}}+2\ln (\sqrt{x-1}+\sqrt{x+1})-2\sqrt{x+1}\bullet$$$$$$$$\Rightarrow$$$$\int \frac{\sqrt{x+1}-1}{\sqrt{x-1}+1}dx=\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{marked}\mathrm{terms}\bullet$$$$\mathrm{please}\mathrm{check}\mathrm{for}\mathrm{typos}$$

Commented by M±th+et£s last updated on 23/Mar/20

$$godblessyouandthankyousirthereis$$$$notypos$$

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