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Question Number 44654 by manish00@gmail.com last updated on 02/Oct/18

$$\int \frac{{\mathbf{e}}^{\mathbf{x}}}{1+{\mathbf{x}}^2}\mathbf{dx}=?$$

Commented by maxmathsup by imad last updated on 02/Oct/18

$$changementx=tan\theta giveI=\int \frac{e^{tan\theta }}{1+{tan}^2\theta }(1+{tan}^2\theta )d\theta$$$$=\int e^{tan\theta }d\theta =\int \left(\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(tan\theta \right)}^n}{n!}\right)d\theta$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\int {tan}^{n}d\theta and{A}_n=\int {tan}^n\theta d\theta canbecalculatedbyrecurrence...$$$$$$

Commented by maxmathsup by imad last updated on 02/Oct/18

$$wehave{e}^x=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}\Rightarrow \int \frac{e^x}{1+x^2}dx=\int \frac{1}{1+x^2}\left(\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}\right)dx$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\int \frac{x^n}{1+x^2}dx=\sum _{n=0}^{\mathrm{\infty }}\frac{A_n}{n!}with{A}_n=\int \frac{x^n}{1+x^2}dx....$$

Commented by maxmathsup by imad last updated on 02/Oct/18

$$ithinkthisintegralcantbecalculatedbyelementaryfunctions...$$

Commented by arvinddayama01@gmail.com last updated on 03/Oct/18

$$Thankusir$$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18

$$isitfeasible...$$

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