The process of decomposing a [[Rational Function]] to make it easier to work with (for example, to [[Integration|Integrate]]). Let $f(x)= \frac{P(x)}{Q(x)}$ be a [[Rational Function]], with $\op{deg}(P(x)) < \op{deg}(Q(x))$. ## Method 1 $Q(x)$ is a product of distinct [[Linear]] [[Factor|Factors]]. $\huge \begin{align} \let N &= \deg(Q(x)) \\ \\ Q(x) &= (a_{1}x+b_{1})(a_{2}x+b_{2})\cdots(a_{N}x+b_{N})\\ &= F_\ba{Q,1} +F_\ba{Q,2} + \cdots F_\ba{Q,N} \end{align} $ $P(x)$ is some [[Polynomial]]: $\huge \begin{align} P(x)=c_{0} + c_{1}x^1 + \cdots \end{align} $ We can write the rational function as: $\huge \begin{align} \frac{P(x)}{Q(x)} &= \frac{A_{1}}{(a_{1}x+b_{1})} + \frac{A_{2}}{(a_{2}x+b_{2})} + \cdots \frac{A_{N}}{(a_{N}x+b_{N})} + \\ &= \frac{A_{1}}{F_\ba{Q,1}}+ \frac{A_{2}}{F_\ba{Q,2}}+\cdots \frac{A_{N}}{F_\ba{Q,N}} \end{align}$ for some constants $A_1, \dots, A_N$. $ \huge \begin{align} \\ \frac{P(x)}{Q(x)} &= \frac{ \sum_{n}^{N}{ A_{n} \prod_{i=1}^{N - 1}{F_\ba{Q , 1+\pa{\pa{n+i}\, \op{mod}\, N} }} }}{ F_\ba{Q,1}+ F_\ba{Q,2}+\cdots F_\ba{Q,N} }\\ P(x)&= \sum_{n}^N A_{n}F_\ba{Q,n} \end{align} $ >[!example] >$\huge \begin{align} > >\frac{x+4}{3x^2-2x-1} &= > >\frac{x+4}{\pa{3x+1}\pa{x-1}} >\\ >&= >\frac{A}{3x+1} + >\frac{B}{x-1} \\ >&= >\frac{A(x-1) + B(3x+1)}{(3x+1)(x-1)} \\ >\\ >\frac{\color{pink}x+4}{\pa{3x+1}\pa{x-1}} &= >\frac{\color{pink}A(x-1) + B(3x+1)}{(3x+1)(x-1)} \\ > >x+4 &=A(x- 1)+B(3x+1)\\ > &= Ax-A+3Bx+B\\ >x+4&=(A+3B)x+B-A \\ > >&\therefore >\\ > >1&=A+3B\\ >4 &=B-A\\ >\end{align}$ > >You can then solve the system of linear equations: >$\huge \begin{align} >B &=4+A \\ >\\ >1&=A+3(4+A) \\ >1&=A+12+3A \\ >-11 &=4A \\ >A &= -\frac{11}{4} \\ >\\ >B &= 4-\frac{11}{4} >\\&= \frac{5}{4} >\\ >\end{align}$ >$\huge \begin{align} > >\frac{x+4}{3x^2-2x-1} &= >\boxed{ >\frac{ -\frac{11}{4} }{3x+1} + >\frac{ \frac{5}{4} }{x-1}} \\ >\end{align}$