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Question Number 36799 by abdo.msup.com last updated on 05/Jun/18

$$find{\int }_0^{\mathrm{\infty }}{e}^{t}ln(1+e^{-2t})dt.$$

Commented by math khazana by abdo last updated on 11/Jun/18

$$letI={\int }_0^{\mathrm{\infty }}{e}^{t}ln(1+e^{-2t})dtbyparts$$$$u^{{}^{\prime }}=e^{t}andv=ln(1+e^{-2t})\Rightarrow$$$$I={\left[e^{t}ln(1+e^{-2t})\right]}_0^{+\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}{e}^t\frac{-2e^{-2t}}{1+e^{-2t}}dt$$$$=-ln\left(2\right)+2{\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{1+e^{-2t}}dtbut$$$${\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{1+e^{-2t}}dt{=}_{e^t=x}{\int }_1^{+\mathrm{\infty }}\frac{1}{x(1+\frac{1}{x^2})}\frac{dx}{x}$$$$={\int }_1^{+\mathrm{\infty }}\frac{dx}{x^2+1}={\left[arctanx\right]}_1^{+\mathrm{\infty }}=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}\Rightarrow$$$$★I=-ln\left(2\right)+\frac{\pi }{2}★$$

Answered by MJS last updated on 06/Jun/18

$$\int {\mathrm{e}}^t\ln (1+{\mathrm{e}}^{-2t})dt=$$$$\left[\begin{array}{c}\int f^{\prime }g=fg-\int {fg}^{\prime }\\ f^{\prime }={\mathrm{e}}^t\Rightarrow f={\mathrm{e}}^t\\ g=\ln (1+{\mathrm{e}}^{-2t})\Rightarrow g^{\prime }=-\frac{{2\mathrm{e}}^{-2t}}{1+{\mathrm{e}}^{-2t}}\end{array}\right]$$$$={\mathrm{e}}^t\ln (1+{\mathrm{e}}^{-2t})-2\int -\frac{{\mathrm{e}}^{-t}}{1+{\mathrm{e}}^{-2t}}dt=$$$$[u=-{\mathrm{e}}^t\to dt=-{\mathrm{e}}^{t}du]$$$$={\mathrm{e}}^t\ln (1+{\mathrm{e}}^{-2t})-2\int \frac{u}{1+u^2}du=$$$$={\mathrm{e}}^t\ln (1+{\mathrm{e}}^{-2t})-2\arctan u=$$$$={\mathrm{e}}^t\ln (1+{\mathrm{e}}^{-2t})-2\arctan -{\mathrm{e}}^t=$$$$={\mathrm{e}}^t\ln (1+{\mathrm{e}}^{-2t})+2\arctan {\mathrm{e}}^t+C$$$$$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^t\ln (1+{\mathrm{e}}^{-2t})dt=\frac{\pi }{2}-\ln 2$$

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