Partial Fractions

$$\frac{5x+3}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}$$ $$\begin{array}{r}5x+3=A(x+2)+B(x+1)\\ 5x+3=(A+B)x+(2A+B)\\ A+B=5,2A+B=3\\ \to A=-2,B=7\end{array}$$ $$\boxed{\frac{5x+3}{(x+1)(x+2)}=\frac{-2}{x+1}+\frac{7}{x+2}}$$

For partial fractions, if you have a repeated factor such as $(x-a{)}^n$ in the denominator, you must have degrees 1 through $n$ in your partial fraction sum.

$$\frac{\cdots }{\cdots (x-a{)}^n}=\cdots +\sum _{i=1}^n\frac{c_i}{(x-a{)}^i}$$

This is really thorough: https://www.mathsisfun.com/algebra/partial-fractions.html

Example

$${\int }_{-2}^2\frac{1}{(x+7)(x^2+9)}dx$$

Use partial fractions on the integrand

$$\begin{array}{rl}\frac{1}{(x+7)(x^2+9)}& =\frac{A}{x+7}+\frac{Bx+C}{x^2+9}\\ 1& =A(x^2+9)+(Bx+C)(x+7)\\ 1& =A(x^2+9)+(Bx+C)(x+7)\\ (0)x^2+(0)x+1& =(A+B)x^2+(7B+C)x+(9A+7C)\\ & \text{3 equations, 3 unknowns}\\ & A=\frac{1}{58},B=-\frac{1}{58},C=\frac{7}{58}\end{array}$$

Finish the integral

$$\begin{array}{rl}\frac{1}{58}{\int }_{-2}^2\left(\frac{1}{x+7}+\frac{7-x}{x^2+9}\right)dx& =\\ \frac{1}{58}{\int }_{-2}^2\left(\frac{1}{x+7}+\frac{7}{x^2+9}-\frac{x}{x^2+9}\right)dx& =\\ \frac{1}{58}\left(\int \frac{1}{x+7}dx+\int \frac{7}{x^2+9}dx-\int \frac{x}{x^2+9}dx\right){|}_{-2}^2& =\\ \frac{1}{58}\left(\int \frac{du}{u}+\int \frac{7}{x^2+9}dx-\int \frac{x}{x^2+9}dx+C\right){|}_{-2}^2& =\\ \frac{1}{58}\left(\ln (x+7)+\int \frac{7}{x^2+9}dx-\int \frac{x}{x^2+9}dx+C\right){|}_{-2}^2& =\\ \frac{1}{58}\left(\ln (x+7)+\int \frac{7}{x^2+9}dx-\frac{1}{2}\int \frac{du}{u}+C\right){|}_{-2}^2& =\\ \frac{1}{58}\left(\ln (x+7)+\int \frac{7}{x^2+9}dx-\frac{1}{2}\ln (x^2+9)+C\right){|}_{-2}^2& =\\ \frac{1}{58}\left(\ln (x+7)+\frac{7}{9}{\tan }^{-1}\left(\frac{x}{3}\right)-\frac{1}{2}\ln (x^2+9)+C\right){|}_{-2}^2& =\\ & \text{and so on}\end{array}$$