$$$\frac{e^{x}}{- 9 x e^{2} + 16}$$$ 的積分

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

令 $$$u=x - \frac{16}{9 e^{2}}$$$。

則 $$$du=\left(x - \frac{16}{9 e^{2}}\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

該積分變為

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=- \frac{1}{9 e^{2}}$$$ 與 $$$f{\left(u \right)} = \frac{e^{u + \frac{16}{9 e^{2}}}}{u}$$$：

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

重寫被積函數:

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=e^{\frac{16}{9 e^{2}}}$$$ 與 $$$f{\left(u \right)} = \frac{e^{u}}{u}$$$：

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

此積分（指數積分）不存在閉式表示：

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

回顧一下 $$$u=x - \frac{16}{9 e^{2}}$$$：

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

因此，

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

化簡：

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

加上積分常數：

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

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