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Question Number 44695 by manish00@gmail.com last updated on 03/Oct/18

$$\int \frac{{\mathbf{e}}^{\sqrt{\mathbf{t}-1}}}{\mathbf{t}}\mathbf{dt}=?$$

Commented by maxmathsup by imad last updated on 04/Oct/18

$$changement\sqrt{t-1}=xgivet-1=x^2\Rightarrow$$$$\int \frac{e^{\sqrt{t-1}}}{t}dt=\int \frac{e^x}{1+x^2}\left(2x\right)dx=\int \frac{2x{e}^x}{1+x^2}dxbyparts{u}^{{}^{\prime }}=\frac{2x}{1+x^2}andv=e^x$$$$I=e^{x}ln(1+x^2)-\int e^{x}ln(1+x^2)dxbut$$$$\int e^{x}ln(1+x^2)dx=\int \left(\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}\right)ln(1+x^2)dx$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\int x^{n}ln(1+x^2)dx=\sum _{n=0}^{\mathrm{\infty }}\frac{A_n}{n!}with{A}_n=\int x^{n}ln(1+x^2)dx$$$$byparts{A}_n=\frac{1}{n+1}{x}^{n+1}ln(1+x^2)-\int \frac{1}{n+1}{x}^{n+1}\frac{2x}{1+x^2}dx$$$$=\frac{x^{n+1}}{n+1}ln(1+x^2)-\frac{2}{n+1}\int \frac{x^{n+1}}{1+x^2}dx...becontinuedanywaytheintegal$$$$\int e^{x}ln(1+x^2)dxisalculableon[0,1]....$$

Commented by arvinddayama01@gmail.com last updated on 04/Oct/18

$$\mathrm{wrong}$$

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