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# Math formulas:Integrals of rational functions

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### Integrals involving $ax + b$

 $$\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)}, \quad (\text{for } n \ne 1)$$
 $$\int \frac{1}{ax+b}dx = \frac{1}{a}\ln|ax+b|$$
 $$\int x (ax+b)^ndx = \frac{a(n+1)x-b}{a^2(n+1)(n+2)}(ax+b)^{n+1}, \quad (\text{for } n \ne -1, n\ne-2)$$
 $$\int \frac{x}{ax+b}dx = \frac{x}{2} - \frac{b}{a^2}\ln|ax+b|$$
 $$\int \frac{x}{(ax+b)^2}dx = \frac{b}{a^2(ax+b)} - \frac{1}{a^2}\ln|ax+b|$$
 $$\int \frac{x^2}{ax+b} dx = \frac{1}{a^3} \left(\frac{(ax+b)^2}{2}-2b(ax+b)+b^2\ln|ax+b| \right)$$
 $$\int \frac{x^2}{(ax+b)^2} dx = \frac{1}{a^3} \left(ax+b-2\,b\,\ln|ax+b| - \frac{b^2}{ax+b}\right)$$
 $$\int \frac{x^2}{(ax+b)^3} dx = \frac{1}{a^3} \left( \ln|ax+b| + \frac{2b}{ax+b} - \frac{b^2}{2(ax+b)^2} \right)$$
 $$\int \frac{x^2}{(ax+b)^n} dx = \frac{1}{a^3} \left( -\frac{(ax+b)^{3-n}}{n-3} + \frac{2b(a+b)^{2-n}}{n-2} - \frac{b^2(ax+b)^{1-n}}{n-1}\right)$$
 $$\int \frac{1}{x(ax+b)}dx = -\frac{1}{b}\ln\left|\frac{ax+b}{x}\right|$$
 $$\int \frac{1}{x^2(ax+b)^2}dx = -\frac{1}{bx} + \frac{a}{b^2} \ln\left|\frac{ax+b}{x}\right|$$
 $$\int \frac{1}{x^2(ax+b)^2}dx = -a\left(\frac{1}{b^2(ax+b)} + \frac{1}{ab^2x} - \frac{2}{b^3}\ln\left|\frac{ax+b}{x} \right|\right)$$

### Integrals involving $ax^2 + bx + c$

 $$\frac{1}{x^2+a^2} dx = \frac{1}{a}\arctan\frac{x}{a}$$
 $$\frac{1}{x^2 - a^2} dx = \frac{1}{2a} \ln\left| \frac{x-a}{x+a}\right|$$
 \int \frac{1}{ax^2 + bx + c} dx = \left\{ \begin{aligned} & \frac{2}{\sqrt{4ac - b^2}} \arctan \frac{2ax + b}{\sqrt{4ac-b^2}} \quad \text{for } 4ac - b^2 > 0 \\ & \frac{2}{\sqrt{b^2 - 4ac}} \ln \left| \frac{2ax + b - \sqrt{b^2-4ac}}{2ax + b + \sqrt{b^2-4ac}} \right| \quad \text{for } 4ac - b^2 < 0 \\ & -\frac{2}{2ax+b} \quad \text{for } 4ac - b^2 = 0 \end{aligned} \right.
 $$\int \frac{x}{ax^2 + bx +c} dx = \frac{1}{2a} \ln\left|ax^2+bx+c\right| - \frac{b}{2a}\int \frac{dx}{ax^2+bx+c}$$
 $$\int \frac{1}{(ax^2+bx+c)^n} dx = \frac{2ax+b}{(n-1)(4ac-b^2)(ax+bx+c)^{n-1}}+ \frac{2(2n-3)a}{(n-1)(4ac - b^2)}\int\frac{dx}{(ax^2+bx+c)^{n-1}}$$
 $$\int \frac{1}{x(ax^2 + bx + c)} dx = \frac{1}{2c}\ln\left|\frac{x^2}{ax^2+bx+c}\right| - \frac{b}{2c} \int \frac{1}{ax^2+bx+c}dx$$