Integral of $$$\frac{e^{\frac{x^{6}}{2}}}{x}$$$

Find $$$\int \frac{e^{\frac{x^{6}}{2}}}{x}\, dx$$$.

Let $$$u=x^{6}$$$.

Then $$$du=\left(x^{6}\right)^{\prime }dx = 6 x^{5} dx$$$ (steps can be seen »), and we have that $$$x^{5} dx = \frac{du}{6}$$$.

So,

$${\color{red}{\int{\frac{e^{\frac{x^{6}}{2}}}{x} d x}}} = {\color{red}{\int{\frac{e^{\frac{u}{2}}}{6 u} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{6}$$$ and $$$f{\left(u \right)} = \frac{e^{\frac{u}{2}}}{u}$$$:

$${\color{red}{\int{\frac{e^{\frac{u}{2}}}{6 u} d u}}} = {\color{red}{\left(\frac{\int{\frac{e^{\frac{u}{2}}}{u} d u}}{6}\right)}}$$

Let $$$v=\frac{u}{2}$$$.

Then $$$dv=\left(\frac{u}{2}\right)^{\prime }du = \frac{du}{2}$$$ (steps can be seen »), and we have that $$$du = 2 dv$$$.

Therefore,

$$\frac{{\color{red}{\int{\frac{e^{\frac{u}{2}}}{u} d u}}}}{6} = \frac{{\color{red}{\int{\frac{e^{v}}{v} d v}}}}{6}$$

This integral (Exponential Integral) does not have a closed form:

$$\frac{{\color{red}{\int{\frac{e^{v}}{v} d v}}}}{6} = \frac{{\color{red}{\operatorname{Ei}{\left(v \right)}}}}{6}$$

Recall that $$$v=\frac{u}{2}$$$:

$$\frac{\operatorname{Ei}{\left({\color{red}{v}} \right)}}{6} = \frac{\operatorname{Ei}{\left({\color{red}{\left(\frac{u}{2}\right)}} \right)}}{6}$$

Recall that $$$u=x^{6}$$$:

$$\frac{\operatorname{Ei}{\left(\frac{{\color{red}{u}}}{2} \right)}}{6} = \frac{\operatorname{Ei}{\left(\frac{{\color{red}{x^{6}}}}{2} \right)}}{6}$$

Therefore,

$$\int{\frac{e^{\frac{x^{6}}{2}}}{x} d x} = \frac{\operatorname{Ei}{\left(\frac{x^{6}}{2} \right)}}{6}$$

Add the constant of integration:

$$\int{\frac{e^{\frac{x^{6}}{2}}}{x} d x} = \frac{\operatorname{Ei}{\left(\frac{x^{6}}{2} \right)}}{6}+C$$

$$$\int \frac{e^{\frac{x^{6}}{2}}}{x}\, dx = \frac{\operatorname{Ei}{\left(\frac{x^{6}}{2} \right)}}{6} + C$$$A