Nice integral solved by substitution asked by Sourav De in math group

Exercise:

Integrate, $\int{\frac{{{x}^{n}}}{\sum\limits_{i=0}^{n}{\frac{{{x}^{i}}}{i!}}}dx}$

Solution: Let ${{u}_{n}}\left( x \right)=\sum\limits_{i=1}^{n}{\frac{{{x}^{i}}}{i!}}=1+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}...+\frac{{{x}^{n-1}}}{\left( n-1 \right)!}+\frac{{{x}^{n}}}{n!}$

So ${{u}_{n}}'\left( x \right)=1+\frac{{{x}^{2}}}{3}+....+\frac{n{{x}^{n-1}}}{n!}=1+\frac{{{x}^{2}}}{2!}+....+\frac{{{x}^{n-1}}}{\left( n-1 \right)!}\Leftrightarrow u{{'}_{n}}\left( x \right)={{u}_{n}}\left( x \right)+\frac{{{x}^{n}}}{n!}$

So ${{u}_{n}}\left( x \right)-u{{'}_{n}}\left( x \right)=\frac{{{x}^{n}}}{n!}\Leftrightarrow {{x}^{n}}=n!\left( {{u}_{n}}\left( x \right)-u{{'}_{n}}\left( x \right) \right)$

Thus $\int{\frac{{{x}^{n}}}{{{u}_{n}}\left( x \right)}dx}=\int{\frac{n!\left( {{u}_{n}}\left( x \right)-u{{'}_{n}}\left( x \right) \right)}{{{u}_{n}}\left( x \right)}dx}=n!\int{\frac{{{u}_{n}}\left( x \right)}{{{u}_{n}}\left( x \right)}dx-n!\int{\frac{d{{u}_{n}}\left( x \right)}{{{u}_{n}}\left( x \right)}}}$

                       $=n!x-n!\ln \left( {{u}_{n}}\left( x \right) \right)+c=n!\left( x-\ln \left( \sum\limits_{i=1}^{n}{\frac{{{x}^{i}}}{i!}} \right) \right)+c$

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The idea of solution Credit  to Igor Soposki