Integral of $$$\frac{e^{x}}{- 9 x e^{2} + 16}$$$

Find $$$\int \frac{e^{x}}{- 9 x e^{2} + 16}\, dx$$$.

Let $$$u=x - \frac{16}{9 e^{2}}$$$.

Then $$$du=\left(x - \frac{16}{9 e^{2}}\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

Therefore,

$${\color{red}{\int{\frac{e^{x}}{- 9 x e^{2} + 16} d x}}} = {\color{red}{\int{\left(- \frac{e^{u + \frac{16}{9 e^{2}}}}{9 u e^{2}}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{1}{9 e^{2}}$$$ and $$$f{\left(u \right)} = \frac{e^{u + \frac{16}{9 e^{2}}}}{u}$$$:

$${\color{red}{\int{\left(- \frac{e^{u + \frac{16}{9 e^{2}}}}{9 u e^{2}}\right)d u}}} = {\color{red}{\left(- \frac{\int{\frac{e^{u + \frac{16}{9 e^{2}}}}{u} d u}}{9 e^{2}}\right)}}$$

Rewrite the integrand:

$$- \frac{{\color{red}{\int{\frac{e^{u + \frac{16}{9 e^{2}}}}{u} d u}}}}{9 e^{2}} = - \frac{{\color{red}{\int{\frac{e^{u} e^{\frac{16}{9 e^{2}}}}{u} d u}}}}{9 e^{2}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=e^{\frac{16}{9 e^{2}}}$$$ and $$$f{\left(u \right)} = \frac{e^{u}}{u}$$$:

$$- \frac{{\color{red}{\int{\frac{e^{u} e^{\frac{16}{9 e^{2}}}}{u} d u}}}}{9 e^{2}} = - \frac{{\color{red}{e^{\frac{16}{9 e^{2}}} \int{\frac{e^{u}}{u} d u}}}}{9 e^{2}}$$

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

$$- \frac{e^{\frac{16}{9 e^{2}}} {\color{red}{\int{\frac{e^{u}}{u} d u}}}}{9 e^{2}} = - \frac{e^{\frac{16}{9 e^{2}}} {\color{red}{\operatorname{Ei}{\left(u \right)}}}}{9 e^{2}}$$

Recall that $$$u=x - \frac{16}{9 e^{2}}$$$:

$$- \frac{e^{\frac{16}{9 e^{2}}} \operatorname{Ei}{\left({\color{red}{u}} \right)}}{9 e^{2}} = - \frac{e^{\frac{16}{9 e^{2}}} \operatorname{Ei}{\left({\color{red}{\left(x - \frac{16}{9 e^{2}}\right)}} \right)}}{9 e^{2}}$$

Therefore,

$$\int{\frac{e^{x}}{- 9 x e^{2} + 16} d x} = - \frac{e^{\frac{16}{9 e^{2}}} \operatorname{Ei}{\left(x - \frac{16}{9 e^{2}} \right)}}{9 e^{2}}$$

Simplify:

$$\int{\frac{e^{x}}{- 9 x e^{2} + 16} d x} = - \frac{\operatorname{Ei}{\left(x - \frac{16}{9 e^{2}} \right)}}{9 e^{2 - \frac{16}{9 e^{2}}}}$$

Add the constant of integration:

$$\int{\frac{e^{x}}{- 9 x e^{2} + 16} d x} = - \frac{\operatorname{Ei}{\left(x - \frac{16}{9 e^{2}} \right)}}{9 e^{2 - \frac{16}{9 e^{2}}}}+C$$

$$$\int \frac{e^{x}}{- 9 x e^{2} + 16}\, dx = - \frac{\operatorname{Ei}{\left(x - \frac{16}{9 e^{2}} \right)}}{9 e^{2 - \frac{16}{9 e^{2}}}} + C$$$A