Use formula

polynomial 多项式

Basically, there are two types of indefinite integral.

One is polynomial, which composed by $+$ and $-$

Another is composite function, which composed by $\times$ and $\div$

For the first type of integral, we use basic formula to each part of it directly:

$$\begin{align*} &\int \frac{3x^2 - 2x + 1}{x} dx \\ \\ =& \int (3x - 2 + \frac{1}{x}) dx \\ \\ =& \frac{3}{2}x^2 - 2x + \ln{|x|} + C \end{align*}$$

For the second one, oh my god, it's very complicate.

Here's a basic template:

$$\int f(g(x)) \cdot g^\prime(x)dx \\ \\ \int f(g(x)) \cdot dg(x) \\ \\ \int f(u) \cdot du$$

Let's do some exercise:

$$\begin{align*} &\int xe^{-x^2} \cdot dx \\ \\ =& \int e^{-x^2} \cdot x dx &\text{//composite function in front} \\ \\ =& (-\frac{1}{2}) \int e^{-x^2} \cdot (-2)x dx &\text{//find a way to make x = } (-x^2)^\prime = -2x \text{ while keeping equation's balance} \\ \\ =& (-\frac{1}{2}) \int e^{-x^2} \cdot d(-x^2) &\text{//we knew }g^\prime(x)dx = dg(x) \\ \\ =& (-\frac{1}{2}) e^{-x^2} + C &\text{//see } -x^2 \text{ as a part, then apply basic integral formula} \end{align*}$$

$$\begin{align*} \\ \\ \\ &\int \cos(1-3x) \cdot dx \\ \\ =&\int \cos(1-3x) \cdot 1dx &\text{//don't forget 1} \\ \\ =& (-\frac{1}{3})\int \cos(1-3x) \cdot (-3)1dx &\text{//find a way to make 1 to -3 while keeping equation's balance} \\ \\ =& (-\frac{1}{3})\int \cos(1-3x) \cdot d(1-3x) &\text{//} \frac{dy}{dx} dx = dy \\ \\ =& (-\frac{1}{3}) \sin(1-3x) + C &\text{//see 1-3x as a part, then use basic formula} \\ \\ \\ \end{align*}$$

From these exercises, we know a criticle principle is: $x$ of $dx$ is the original function, it can be just a $x$, it also can be, for example, $x^2$.

$$\begin{align*} \int (x) dx^2 = \int ((x^2)^\prime \cdot x) dx = \int (2x \cdot x) dx = \int (2x^2) dx \end{align*}$$

$$\begin{align*} \\ &\int (1 \cdot x) dx &\text{where's 1 come from? } x^\prime = 1 \end{align*}$$