Indefinite integral $\int\frac{e^x}{x(1+\log(x))}dx$

Question

How to integrate this integral

$$\int\frac{e^x}{x(1+\log(x))}dx$$

My attempt: I try some subtitutions, $e^x=u$$\hspace{0.2cm}$ and $\hspace{0.2cm}$$1+\log(x)=u$ but these are not helpful.Please help me to solve this.

Answer 1

Let $u=\log x$ ,

Then $x=e^u$

$dx=e^u~du$

$\therefore\int\dfrac{e^x}{x(1+\log x)}dx=\int\dfrac{e^{e^u}}{1+u}du$

Let $v=\log(1+u)$ ,

Then $u=e^v-1$

$du=e^v~dv$

$\therefore\int\dfrac{e^{e^u}}{1+u}du$

$=\int e^{e^{e^v-1}}~dv$

$=\int\left(1+\sum\limits_{m=1}^\infty\dfrac{e^{me^v-m}}{m!}\right)dv$

$=\int\left(1+\sum\limits_{m=1}^\infty\dfrac{e^{-m}}{m!}+\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\dfrac{m^ne^{-m}e^{nv}}{m!n!}\right)dv$

$=\int\left(\sum\limits_{m=0}^\infty\dfrac{e^{-m}}{m!}+\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\dfrac{m^ne^{-m}e^{nv}}{m!n!}\right)dv$

$=\sum\limits_{m=0}^\infty\dfrac{e^{-m}v}{m!}+\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\dfrac{m^ne^{-m}e^{nv}}{m!n!n}+C$

$=\sum\limits_{m=0}^\infty\dfrac{e^{-m}\log(1+\log x)}{m!}+\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\dfrac{m^ne^{-m}(1+\log x)^n}{m!n!n}+C$