不定积分

Indefinite Integral

在区间 $I$ 上，可导函数 $F (x)$ 的导函数为 $f (x)$ ，则称 $F (x)$ 为 $f (x)$ 在区间 $I$ 上的原函数
原函数存在定理：连续函数一定有原函数

\begin{aligned} F^{'} (x) = f (x) & \leftrightarrow d F (x) = f (x) d x \int f (x) d x = F (x) + C \end{aligned}

\begin{aligned} \int [f (x) + g (x)] d x & = \int f (x) d x + \int g (x) d x \\ \int k f (x) d x & = k \int f (x) d x \end{aligned}

\begin{aligned} \int k d x = k x + C \\ \int x^{μ} d x = \frac{1}{μ + 1} x^{μ + 1} + C \\ \int e^{x} d x = e^{x} + C \\ \int a^{x} d x = \frac{1}{\ln a} a^{x} + C \\ \int \frac{1}{x} d x = \ln | x | + C \\ \int \frac{1}{1 + x^{2}} d x = \arctan x + C \\ \int \frac{1}{\sqrt{1 - x^{2}}} d x = \arcsin x + C \\ \int \cos x d x = \sin x + C \\ \int \sin x d x = - \cos x + C \\ \int \frac{1}{\cos^{2} x} d x = \int \sec^{2} x d x = \tan x + C \\ \int \frac{1}{\sin^{2} x} d x = \int \csc^{2} x d x = - \cot x + C \\ \int \sec x \tan x d x = \sec x + C \\ \int \csc x \cot x d x = - \csc x + C \end{aligned}

\begin{aligned} \int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \arctan \frac{x}{a} + C \\ \int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \arcsin \frac{x}{a} + C \\ \int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} \ln | \frac{x - a}{x + a} | + C \\ \int \tan x d x = - \ln | \cos x | + C \\ \int \cot x d x = \ln | \sin x | + C \\ \int \sec x d x = \ln | \sec x + \tan x | + C \\ \int \csc x d x = \ln | \csc x - \cot x | + C \end{aligned}

\begin{aligned} \int \frac{d x}{\sqrt{x^{2} + a^{2}}} = \ln (x + \sqrt{x^{2} + a^{2}}) + C \\ \int \frac{d x}{\sqrt{x^{2} - a^{2}}} = \ln | x + \sqrt{x^{2} - a^{2}} | + C_{1} \end{aligned}