Partial fraction expansion, also know as Fractional Decomposition, is a method of expanding an algebraic fraction into the sum of several expressions.


For partial fraction expansion to be possible, the denominator must be of a higher degree than the numerator. If it is, it must be factored. For example:

$ \frac{x+2}{x^2+5x+4}=\frac{x+2}{(x+4)(x+1)} $

Now we can make the numerators equal to $ A,B $ and cross multiplying.

$ \frac{x+2}{(x+4)(x+1)}=\frac{A}{(x+4)}+\frac{B}{(x+1)}=\frac{A(x+1)+B(x+4)}{(x+4)(x+1)} $

We can now multiply by the denominator.

$ x+2=A(x+1)+B(x+4) $

$ A,B $ can be solved for setting $ x $ equal to a number that will cancel one term out.

$ \begin{align}&x=-4 \\&(-4)+2=A\big((-4)+1\big)+B\big((-4)+4\big) \\&-2=A(-3)+B(0)=-3A \\&-2=-3A \\&A=\frac23 \\\\&x=-1 \\&(-1)+2=A\big((-1)+1\big)+B\big((-1)+4\big) \\&1=A(0)+B(3)=3B \\&1=3B \\&B=\frac13\end{align} $

We now know that

$ \frac{x+2}{(x+4)(x+1)}=\frac{\frac{2}{3}}{(x+4)}+\frac{\frac{1}{3}}{(x+1)}=\frac{2}{3(x+4)}+\frac{1}{3(x+1)} $

which can be proved by cross multiplication.


The main use of partial fraction expansion is in integral calculus, where it can be used to find an antiderivative. Since we know that

$ \int\big(f(x)+g(x)\big)dx=\int f(x)dx+\int g(x)dx $

we can use partial fraction expansion to solve seemingly difficult integrals. For example, we just found that

$ \frac{x+2}{(x+4)(x+1)}=\frac{2}{3(x+4)}+\frac{1}{3(x+1)} $

This means that

$ \begin{align}\int\dfrac{x+2}{(x+4)(x+1)}dx&=\int\left(\frac{2}{3(x+4)}+\frac{1}{3(x+1)}\right)dx\\&=\int\frac{2}{3(x+4)}dx+\int\frac{dx}{3(x+1)}\end{align} $

This can easily be solved to find that

$ \int\dfrac{x+2}{(x+4)(x+1)}dx=\frac{2\ln\big(|x+4|\big)+\ln\big(|x+1|\big)}{3}+C $


Integration by Fractional Decomposition-0

Integration by Fractional Decomposition-0

A walkthough of an integration by partial fractions problem.

Community content is available under CC-BY-SA unless otherwise noted.