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Question Number 92156 by abdomathmax last updated on 05/May/20

$$find{\int }_0^1{xe}^{-x^2}ln(1+x)dx$$

Commented by mathmax by abdo last updated on 08/May/20

$$I={\int }_0^{1}x{e}^{-x^2}ln(1+x)dxbypsrts$$$$I={[-\frac{1}{2}{e}^{-x^2}ln(1+x)]}_0^1+\frac{1}{2}{\int }_0^1\frac{e^{-x^2}}{1+x}dx$$$$=-\frac{1}{2}{e}^{-1}ln\left(2\right)+\frac{1}{2}{\int }_0^1\frac{e^{-x^2}}{1+x}dx$$$${\int }_0^1\frac{e^{-x^2}}{1+x}dx{=}_{1+x=t}{\int }_1^2\frac{e^{-{(t-1)}^2}}{t}dt={\int }_1^2\frac{e^{-t^2+2t-1}}{t}dt$$$$=e^{-1}{\int }_1^2\frac{e^{-t^2+2t}}{t}dt=e^{-1}{\int }_1^2\frac{1}{t}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-t^2+2t)}^n}{n!}dt$$$$=e^{-1}{\int }_1^2(\frac{1}{t}+\frac{1}{t}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-t^2+2t)}^n}{n!})dt$$$$=e^{-1}ln\left(2\right)+e^{-1}\sum _{n=1}^{\mathrm{\infty }}{\int }_1^2\frac{t^{n-1}{(2-t)}^n}{n!}dt\Rightarrow$$$$I=\frac{1}{2e}\sum _{n=1}^{\mathrm{\infty }}\frac{A_n}{n!}with{A}_n={\int }_1^2{t}^{n-1}{(2-t)}^{n}dt....becontinued...$$

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