指数积分表

这里列举了一些有关指数函数的不定积分公式，以下例子中均省略积分常数，且限定积分和结果均在被积函数定义域的内点的邻域中。大部分以 $\text{e}^{ax}$ 为例说明， $a \neq 0.$ （注意到一般的指数函数有关系式 $u^x = \text{e}^{x\ln u}$ ）。

指数函数和幂函数的乘积[]

$\int \text{e}^{ax} \mathrm{d}x = \dfrac{1}{a} \text{e}^{ax}.$
$\int x \text{e}^{ax} \mathrm{d}x = \dfrac{1}{a^2} (ax - 1) \text{e}^{ax}.$
$\int x^n \text{e}^{ax} \mathrm{d}x = \dfrac{1}{a} x^n \text{e}^{ax} - \dfrac{n}{a} \int x^{n-1} \text{e}^{ax} \mathrm{d}x.$
$\int x^{-n} \text{e}^{ax} \mathrm{d}x = - \dfrac{1}{(n-1)} x^{-n+1} \text{e}^{ax} + \dfrac{c}{n-1} \int x^{-n+1} \text{e}^{ax} \mathrm{d}x, \quad n \ne -1.$
$\int x \text{e}^{ax^2} \mathrm{d}x = \dfrac{1}{2a} \text{e}^{ax^2}.$
$\int x^{2n-1} \text{e}^{x^n} \mathrm{d}x = \dfrac{1}{n} (x^n-1) \text{e}^{x^n}, n \ne 0.$
$\int x^{mn-1} \text{e}^{ax^n} \mathrm{d}x = \dfrac{1}{na} x^{(m-1)n} \text{e}^{ax^n} - \dfrac{m-1}{a} \int x^{(m-1)n-1} \text{e}^{ax^n} \mathrm{d}x, m > 0, n \ne 0.$

$I_1 := \int \text{e}^{ax} \cos bx \mathrm{d}x, I_2 := \int \text{e}^{ax} \sin bx \mathrm{d}x.$
${\displaystyle I_1 + \text{i} I_2 = \int \text{e}^{ax} (\cos bx + \text{i} \sin bx) \mathrm{d}x = \int \text{e}^{(a+\text{i}b)x} \mathrm{d}x = \dfrac{1}{a+\text{i}b} \text{e}^{a+\text{i}b}.}$

${\displaystyle I_1 - \text{i} I_2 = \int \text{e}^{ax} (\cos bx - \text{i} \sin bx) \mathrm{d}x = \int \text{e}^{(a-\text{i}b)x} \mathrm{d}x = \dfrac{1}{a-\text{i}b} \text{e}^{a-\text{i}b}.}$

$I_1 = \dfrac{\text{e}^{ax}}{a^2+b^2} (a\cos bx + b\sin bx), I_2 = \dfrac{\text{e}^{ax}}{a^2+b^2} (a\sin bx - b\cos bx).$
$\int \text{e}^{ax} \sin^n x \mathrm{d}x = \dfrac{e^{ax} \sin^{n-1}x}{a^2+n^2}(a\sin x - n \cos x) + \dfrac{n(n-1)}{a^2+n^2} \int \text{e}^{ax} \sin^{n-2} x \mathrm{d}x.$
$\int \text{e}^{ax} \cos^n x \mathrm{d}x = \dfrac{e^{ax} \cos^{n-1}x}{a^2+n^2}(a\cos x + n \sin x) + \dfrac{n(n-1)}{a^2+n^2} \int \text{e}^{ax} \cos^{n-2} x \mathrm{d}x.$
$I_n := \int x^n \text{e}^{ax} \cos bx \mathrm{d}x, \quad J_n := \int x^n \text{e}^{ax} \sin bx \mathrm{d}x.$
$I_n = \dfrac{x^n \text{e}^{ax}}{a^2+b^2} (a\cos bx + b\sin bx) - \dfrac{n}{a^2+b^2} (aI_{n-1}+bJ_{n-1}), \quad (n > 1).$

$J_n = \dfrac{x^n \text{e}^{ax}}{a^2+b^2} (a\sin bx - b\cos bx) + \dfrac{n}{a^2+b^2} (bJ_{n-1}-aI_{n-1}), \quad (n > 1).$

一般来说，遇到含有指数函数的有理分式，可以整体换元 $t = \text{e}^x$ 或双曲换元或配对积分法。

$\int \dfrac{1}{a\text{e}^x + b\text{e}^{-x}}\mathrm{d}x = \dfrac{1}{\sqrt{ab}} \arctan \left( \sqrt{\dfrac{a}{b}} \text{e}^x \right).$
$\int \dfrac{\text{e}^x}{a\text{e}^x + b\text{e}^{-x}}\mathrm{d}x = \dfrac{1}{2a} \left( x + \ln|a\text{e}^x + b\text{e}^{-x}| \right).$
${\displaystyle \int \dfrac{1}{a\text{e}^x + b\text{e}^{-x} + c}\mathrm{d}x = \begin{cases} \dfrac{1}{\sqrt{c^2-4ab}} \ln \left| \dfrac{2a\text{e}^x + c - \sqrt{c^2-4ab}}{2a\text{e}^x + c + \sqrt{c^2-4ab}} \right|, & c^2 > 4ab, \\ -\dfrac{1}{a\text{e}^x + c}, & c^2 = 4ab, \\ \dfrac{2}{\sqrt{4ab-c^2}} \arctan \dfrac{2a\text{e}^x+c}{\sqrt{4ab-c^2}}, & c^2 < 4ab. \end{cases}}$
设 $J_n := \int \dfrac{\mathrm{d}x}{(a\text{e}^x+b\text{e}^{-x})^n}, n \in \N$ $J_{n}:=\int {\dfrac {\mathrm {d} x}{(a{\text{e}}^{x}+b{\text{e}}^{-x})^{n}}},n\in \mathbb {N}$ ，那么
1. $J_n = \dfrac{1}{4ab(n-1)} \left( (n-2)J_{n-2} + \dfrac{a\text{e}^x - b\text{e}^{-x}}{(a\text{e}^x+b\text{e}^{-x})^{n-1}} \right), n > 1.$
2. $\int \dfrac{\text{e}^x}{(a\text{e}^x+b\text{e}^{-x})^n}\mathrm{d}x = \dfrac{1}{2a} \left( J_{n-1} - \dfrac{1}{n-1} \dfrac{1}{(a\text{e}^x+b\text{e}^{-x})^{n-1}} \right), n > 1.$
3. $\int \dfrac{\text{e}^{-x}}{(a\text{e}^x+b\text{e}^{-x})^n}\mathrm{d}x = \dfrac{1}{2b} \left( J_{n-1} + \dfrac{1}{n-1} \dfrac{1}{(a\text{e}^x+b\text{e}^{-x})^{n-1}} \right), n > 1.$