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Question Number 180680 by cortano1 last updated on 15/Nov/22

Answered by ARUNG_Brandon_MBU last updated on 16/Nov/22

$$I={\int }_0^1\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx,x=\sinh \theta$$$$={\int }_0^{\ln (1+\sqrt{2})}\frac{\sinh \theta -1+\cosh \theta }{\sinh \theta +1+\cosh \theta }\left(\cosh \theta d\theta \right)$$$$={\int }_0^{\ln (1+\sqrt{2})}\frac{e^{\theta }-1}{e^{\theta }+1}\cdot \frac{e^{2\theta }+1}{2e^{\theta }}d\theta =\frac{1}{2}{\int }_1^{1+\sqrt{2}}\frac{(t-1)(t^2+1)}{t^2+t}\left(\frac{dt}{t}\right)$$$$=\frac{1}{2}{\int }_1^{1+\sqrt{2}}\frac{t^3-t^2+t-1}{t^3+t^2}dt=\frac{1}{2}{\int }_1^{1+\sqrt{2}}(1-\frac{2t^2-t+1}{t^3+t^2})dt$$$$=\frac{\sqrt{2}}{2}-\frac{1}{2}{\int }_1^{1+\sqrt{2}}\frac{2t^2-t+1}{t^2(t+1)}dt=\frac{\sqrt{2}}{2}-\frac{1}{2}{\int }_1^{1+\sqrt{2}}(\frac{4}{t+1}-\frac{2t-1}{t^2})dt$$$$=\frac{\sqrt{2}}{2}-\frac{1}{2}{[4\ln (t+1)-2\ln \left(t\right)-\frac{1}{t}]}_1^{1+\sqrt{2}}$$$$=\frac{\sqrt{2}}{2}-2\ln \left(\frac{2+\sqrt{2}}{2}\right)+\ln (1+\sqrt{2})+\frac{1/2}{1+\sqrt{2}}-\frac{1}{2}$$$$=\frac{\sqrt{2}-1}{2}-2\ln (2+\sqrt{2})+\ln 4+\ln (1+\sqrt{2})+\frac{\sqrt{2}-1}{2}$$$$=\sqrt{2}-1+\ln 4-2\ln (2+\sqrt{2})+\ln (1+\sqrt{2})$$

Commented by MJS_new last updated on 16/Nov/22

$$\ln 4-2\ln (2+\sqrt{2})+\ln (1+\sqrt{2})=$$$$=\ln \left(\frac{4(1+\sqrt{2})}{{(2+\sqrt{2})}^2}\right)=-\ln \frac{1+\sqrt{2}}{2}$$$$\Rightarrow \mathrm{our}\mathrm{answers}\mathrm{are}\mathrm{equal}$$

Commented by cortano1 last updated on 16/Nov/22

$$\mathrm{yes}..\mathrm{thx}$$

Answered by MJS_new last updated on 16/Nov/22

$$\mathrm{without}\mathrm{substitution}:$$$$\int \frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx=$$$$=\int \frac{-1+\sqrt{x^2+1}}{x}dx=$$$$=\int (\frac{1+\sqrt{x^2+1}}{x}-\frac{2}{x})dx=$$$$=\int (\frac{x}{-1+\sqrt{x^2+1}}-\frac{2}{x})dx=$$$$=\int (\frac{x}{-1+\sqrt{x^2+1}}-\frac{x}{\sqrt{x^2+1}}+\frac{x}{\sqrt{x^2+1}}-\frac{2}{x})dx=$$$$=\int (\frac{x}{\sqrt{x^2+1}}\times \frac{1}{-1+\sqrt{x^2+1}}+\frac{x}{\sqrt{x^2+1}}-\frac{2}{x})dx=$$$$=\int \frac{d[-1+\sqrt{x^2+1}]}{-1+\sqrt{x^2+1}}+\int d\left[\sqrt{x^2+1}\right]-2\int \frac{dx}{x}=$$$$=\ln (-1+\sqrt{x^2+1})+\sqrt{x^2+1}-2\ln x+C$$$$\Rightarrow$$$$\mathrm{answer}\mathrm{is}-1+\sqrt{2}-\ln \frac{1+\sqrt{2}}{2}$$

Commented by cortano1 last updated on 16/Nov/22

$$\mathrm{yes}\mathrm{thx}$$

Answered by MJS_new last updated on 16/Nov/22

$$\int \frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx=$$$$[t=x+\sqrt{x^2+1}\to dx=\frac{t^2+1}{2t^2}dt]$$$$=\frac{1}{2}\int \frac{(t-1)(t^2+1)}{t^2(t+1)}dt=$$$$=\int (\frac{1}{2}+\frac{1}{t}-\frac{2}{t+1}-\frac{1}{2t^2})dt=$$$$=\frac{1}{2}t+\ln t-2\ln (t+1)+\frac{1}{2t}=$$$$=\frac{t^2+1}{2t}+\ln \frac{t}{{(t+1)}^2}=$$$$=\sqrt{x^2+1}+\ln \frac{-1+\sqrt{x^2+1}}{x^2}+C$$

Answered by Gamil last updated on 16/Nov/22

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