Partial fraction decomposition

Differentials and integrals: Integration techniques

Partial fraction decomposition

For rational functions, an effective approach is sometimes to rewrite a rational function as a sum of simpler rational functions. This method is known as partial fraction decomposition. A few examples illustrate this approach.

Example 1

$\int \frac{1}{y(1-y)}\,\dd y$

In partial fraction decomposition one first tries to find $A$ and $B$ such that

$\frac{1}{y(1-y)}=\frac{A}{y}+\frac{B}{1-y}.$ In order to find $A$ and $B$ , we work out the right-hand side by bringing the terms under a common denominator:

$\begin{aligned}\frac{A}{y}+\frac{B}{1-y}&=\frac{A(1-y)}{y(1-y)}+\frac{B y}{y(1-y)}\\[0.25cm] &=\frac{A(1-y)+B y}{y(1-y)}\\[0.25cm] &=\frac{(B-A)y+A}{y(1-y)}\end{aligned}$ Thus, we get the equations $B-A=0$ and $A=1$ . So $A=1$ and $B=1$ , that is,

$\frac{1}{y(1-y)}=\frac{1}{y}+\frac{1}{1-y}$ Then the following reasoning is correct:

$\begin{aligned}\int\frac{1}{y(1-y)}\,\dd y&=\int\frac{1}{y}\,\dd y+\int\frac{1}{1-y}\dd y\\[0.25cm] &=\ln\bigl(|y|\bigr)-\ln\bigl(|1-y|\bigr)+c\end{aligned}$

Not all denominators of rational functions can be split to some factors of degree 1 as above. Some polynomials have factors of degree 2 that cannot be further split into real factors. See the following examples to find out how to use partial fraction decomposition in such case.

Example 2

$\int \frac{1}{x^3+4x}\,\dd x$

In partial fraction decomposition one first tries to find $A$ $A$ , $B$ and $C$ such that

$\begin{aligned}\frac{1}{x^3+4x} &= \frac{1}{x(x^2+4)}\\[0.25cm] &=\frac{A}{x}+\frac{Bx+C}{x^2+4}, \end{aligned}$ In order to find $A$ , $B$ and $C$ , we work out the right-hand side by bringing the terms under a common denominator:

$\begin{aligned}\frac{x}{x^3+4x} &=\frac{A(x^2+4) + (Bx+C)x}{x^3+4x}\\[0.25cm]&=\frac{(A+B)x^2 +Cx + 4A}{x^3+4x}\end{aligned}$ Thus, we get the equations $A+B = 0$ , $C = 0$ , and $4A = 1$ . So $A=\frac{1}{4}$ , $B=-\frac{1}{4}$ , $C=0$ . We get

$\begin{aligned} \int \frac{1}{x^3+4x}\,\dd x &=\frac{1}{4}\int\frac{1}{x}\dd x -\frac{1}{4} \int \frac{x}{x^2+4}\dd x\\[0.25cm] &=\frac{1}{4}\ln|x| -\frac{1}{8}\ln(x^2+4)+c \end{aligned}$ In the second logarithmic term, we omit absolute value bars because $x^2+4$ is always positive.

Example 3

$\int \frac{2 x^2 - x + 4}{x^3 + 4x} \, \dd x$

The method of partial fraction decomposition goes as follows:

$\begin{aligned}\int \frac{2 x^2 - x + 4}{x^3 + 4x}\, \dd x & = \int \frac{2 x^2 - x + 4}{x ( x^2 + 4)}\,\dd x\\[0.25cm] &= \int\left( \frac{1}{x} + \frac{x-1}{x^2+4}\right) \dd x \\[0.25cm] &= \int \left(\frac{1}{x} + \frac{x}{x^2+4} + \frac{-1}{x^2+4}\right) \dd x \\[0.25cm] &= \ln |x| + \frac{1}{2}\ln |x^2+4| - \frac{1}{2}\arctan (\tfrac{1}{2}x)\end{aligned}$
In the first step, the denominator is factored. Note that $x^2 + 4$ cannot be factored further. The second step is partial fraction decomposition; here, $A, B, C$ are solved from

$\frac{2 x^2 - x + 4}{x ( x^2 + 4)} = \frac{A}{x} + \frac{B x + C}{x^2+4}$
The calculation, which we omit, yields $A = 1, B = 1, C=-1$ . Because $x^2+4$ is a quadratic term, a linear term is in the denominator. Finally, the integral $\int \dfrac{x-1}{x^2+4},dx$ is split into two integrals, namely $\int \dfrac{x}{x^2+4},\dd x$ and $\int \dfrac{-1}{x^2+4}, \dd x$ .

The first of these can be tackled using the substitution $u = x^2 + 4$ : with $\dd u = 2x ,\dd x$ , we get

$\int \frac{x}{x^2+4}\, \dd x = \int \frac{1}{2} \frac{1}{u}\, \dd u = \frac{1}{2}\ln |u| = \frac{1}{2}\ln |x^2+4|$ The second can be simplified using the substitution $x = 2v$ . Then, with $\dd x = 2, \dd v$ , we see

$\begin{aligned} \int \frac{-1}{x^2+4}\, \dd x&= \int 2 \frac{-1}{4 v^2 + 4}\, \dd v\\[0.25cm] &= - \frac{1}{2}\int\frac{1}{v^2 + 1}\, \dd v\\[0.25cm] &= \frac{1}{2}\arctan (v)\\[0.25cm] &= \frac{1}{2}\arctan(\tfrac{1}{2}x)\end{aligned}$

Another situation that can occur is that the denominator can be factorised to factors of degree 1 or 2 that occur more than once. See below for worked-out examples in that situation.

Example 4 The following is true:

$\frac{1}{x(x+1)^2} = \frac{1}{x} + \frac{-x - \frac{1}{2}}{(x+1)^2}$
To find this, write $\dfrac{1}{x(x+1)^2} = \dfrac{A}{x} + \dfrac{Bx + C}{(x+1)^2}$ and solve for $A, B, C$ . Note that there remains a term with $(x+1)^2$ in the denominator, where the power of the numerator is one less.

Example 5 Consider the function

$f(x) = \frac{x^4-1}{x(x+1)^2}$ The numerator contains a term of degree four (as the highest power), which is higher than the third-degree term that appears as the highest power in the denominator when factoring. To find the antiderivative of $f(x)$ , we can first perform long division of the numerator by the denominator, rewriting $f(x)$ as a polynomial plus a remainder term, which is a rational function with denominator $x(x+1)^2 = x^3 + 2x^2 + x$ and a numerator with terms of degree less than 3. Specifically,

$\begin{aligned}x^4 - 1 &= x \left( x^3 + 2x^2 + x \right) - 2x^3 - x^2 &= \\[0.25cm] &= x \left( x^3 + 2x^2 + x\right) - 2 \left( x^3 + 2x^2 + x \right) + 3 x^2 +2 x\end{aligned}$ From this, we have:

$\begin{aligned}\frac{x^4-1}{x (x+1)^2} &= x - 2 + \frac{ 3 x^2 +2 x}{x (x+1)^2}\\[0.25cm] &= x - 2 + \frac{3x + 2}{(x+1)^2}\\[0.25cm] &= x - 2 + \frac{3(x + 1) -1}{(x+1)^2}\\[0.25cm] &= x - 2 + \frac{3}{(x+1)}-\frac{1}{(x+1)^2}\end{aligned}$

Example 6

$\int \frac{1}{x^3+2x^2}\dd x.$

We can now split the fraction as follows:

$\begin{aligned} \frac{1}{x^3+2x^2} &= \frac{1}{x^2(x+2)}\\[0.25cm] & = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+2}, \end{aligned}$ and we find the unknowns $A$ , $B$ , and $C$ by working out the right-hand side by bringing the terms under a common denominator:

$\begin{aligned} \frac{1}{x^3+2x^2} &= \frac{Ax(x+2) + B(x+2) + Cx^2}{x^3 + 2x^2}\\[0.25cm] &\frac{(A+C)x^2 + (2A+B)x + 2B}{x^3+2x^2}. \end{aligned}$ Thus, we get the equations $A+C = 0$ , $2A + B = 0$ and $2B = 1$ . That gives $A=-\frac{1}{4}$ , $B = \frac{1}{2}$ , and $C=\frac{1}{4}$ . We get

$\begin{aligned} \int\frac{1}{x^3+2x^2}\dd x &=-\frac{1}{4}\int\frac{1}{x}\dd x + \frac{1}{2}\int\frac{1}{x^2}\dd x + \frac{1}{4}\int\frac{1}{x+2}\dd x\\[0.25cm]& = -\frac{1}{4}\ln|x| - \frac{1}{2x} + \frac{1}{4}\ln|x+2| + c\end{aligned}$