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Integration using partial fractions

How do you integrate an expression when there is an algebraic expression in the numerator and denominator of a fraction? Integrating using partial fractions helps you to solve this problem. Use this resource to learn how to integrate using partial fractions.

Sometimes a complex function may be integrated by breaking it up into partial fractions. The idea is that each partial fraction is an easier integral than the original.

For example:
\[\int\frac{1}{x^{2}+5x+6}dx=\int\frac{1}{x+2}dx-\int\frac{1}{x+3}dx\]

The integrals on the right are much simpler than that on the left.

Before you dive into this resource, make sure you are confident with finding partial fractions.

Example 1 – integrating using partial fractions

Find the integral of \(\dfrac{x-5}{x^{2}+2x-3}\) with respect to \(x\).

First, find the partial fractions. Since \(x^{2}+2x-3=(x+3)(x-1)\), we are dealing with the case where the denominator has two distinct linear factors. Therefore, we factorise as follows:
\[\begin{align*} \frac{x-5}{x^{2}+2x-3} & = \frac{x-5}{(x+3)(x-1)}\\
& = \frac{A}{x+3}+\frac{B}{x-1}
\end{align*}\]

Now, we have to determine the constants \(A\) and \(B\). First, add the partial fractions by finding a common denominator. In this case, it is \((x+3)(x-1)\).
\[\begin{align*} \frac{A}{x+3}+\frac{B}{x-1} & = \frac{A}{(x+3)}\times\frac{(x-1)}{(x-1)}+\frac{B}{(x-1)}\times\frac{(x+3)}{(x+3)}\\
& =\frac{A(x-1)+B(x+3)}{(x+3)(x-1)}\\
& =\frac{Ax-A+Bx+3B}{(x+3)(x-1)}\\
& =\frac{Ax+Bx-A+3B}{x^{2}+2x-3}\\
& =\frac{(A+B)x-A+3B}{x^{2}+2x-3}
\end{align*}\]

Equating the numerators:
\[x-5=(A+B)x-A+3B\]

where \(A+B\) is the \(x\) term and \(-A+3B\) is the constant. On the left, the coefficient of \(x\) is \(1\) and on the right, the coefficient is \(A+B\). Similarly, the constant term on the left is \(-5\) and on the right, it is \(-A+3B\).

This gives us the following simultaneous equations:
\[\begin{align*} A+B & = 1 \\
-A+3B & = -5
\end{align*}\]

These equations are easily solved by elimination. Adding the equations, we get:
\[\begin{align*} 4B & = -4\\
B & = -1
\end{align*}\]

Substituting for \(B\) the first equation gives:
\[\begin{align*} A-1 & = 1\\
A & = 2
\end{align*}\]

Finally, we have:
\[\frac{x-5}{x^{2}+2x-3}=\frac{2}{x+3}-\frac{1}{x-1}\]

We can now evaluate the integral:
\[\begin{align*} \int\frac{x-5}{x^{2}+2x-3}dx & = \int\left(\frac{2}{x+3}-\frac{1}{x-1}\right)dx\\
& = \int\frac{2}{x+3}dx-\int\frac{1}{x-1}dx\\
& = 2\ln|x+3|-\ln|x-1|+c,\,c\in\mathbb{R}\\
& = \ln(x+3)^{2}-\ln|x-1|+c\\
& = \ln\frac{(x+3)^{2}}{|x-1|}+c
\end{align*}\]

Find the integral of \(\dfrac{x^{2}-3x+16}{x^{3}-5x^{2}+x-5}\) with respect to \(x\).

We, first factorise the denominator.
\[\begin{align*} x^{3}-5x^{2}+x-5 & = x^{2}(x-5)+x-5\\
& = (x^{2}+1)(x-5)
\end{align*}\]

In this case, we have a quadratic factor and a linear factor. Therefore, we factorise as follows:
\[\begin{align*} \frac{x^{2}-3x+16}{x^{3}-5x^{2}+x-5} & = \frac{Ax+B}{x^{2}+1}+\frac{C}{x+5}\\
& = \frac{(Ax+B)(x-5)}{(x^{2}+1)(x-5)}+\frac{C(x^{2}+1)}{(x^{2}+1)(x-5)}\\
& = \frac{Ax^{2}-5Ax+Bx-5B+Cx^{2}+C}{(x^{2}+1)(x+5)}\\
& = (A+C)x^{2}+(-5A+B)x-5B+C
\end{align*}\]

Equating coefficients of \(x^{2}\), \(x\) and the constant gives the three equations:
\[\begin{align*} A+C & = 1\\
-5A+B & = -3 \\
-5B+C & = 16.
\end{align*}\]

To solve them simultaneously, we can first multiply the first equation by \(5\).
\[5A+5C = 5\]

We can add the second equation to this new equation to get: \[\begin{align*} B+5C & = -3+5\\
& = 2\\
5B+25C & = 10
\end{align*}\]

Adding the same equation to the third equation gives:
\[\begin{align*} 26C & = 26\\
C & = 1
\end{align*}\]

Substituting \(C=1\) into the first equation gives:
\[\begin{align*} A+1 & = 1\\
A & = 0
\end{align*}\]

Finally, substituting \(A=0\) into the second equation gives:
\[B = -3\]

Therefore, we have:
\[\frac{x^{2}-3x+16}{x^{3}-5x^{2}+x-5} = \frac{-3}{x^{2}+1}+\frac{1}{x-5}\]

We can now find the integral:
\[\begin{align*} \int\left(\frac{x^{2}-3x+16}{x^{3}-5x^{2}+x-5}\right)dx & = \int\left(\frac{-3}{x^{2}+1}+\frac{1}{x-5}\right)dx\\
& = \int\frac{1}{x-5}dx-3\int\frac{1}{x^{2}+1}dx\\
& = \ln|x-5|-3\tan^{-1}(x)+c,\,c\in\mathbb{R}
\end{align*}\]

Exercise – integrating using partial fractions

Perform the following integrations.

  1. \(\int\dfrac{4x+9}{x^{2}+x-12}dx\)
  2. \(\int\dfrac{21-8x}{x^{2}-x-6}dx\)
  3. \(\int\dfrac{5x^{2}-5x+2}{(x+1)(x-1)^{2}}dx\)

  1. \(\ln(\left|3-x\right|^{3}\left|x+4\right|)+c\)
  2. \(-\dfrac{3}{5}\ln\left|3-x\right|-\dfrac{37}{5}\ln\left|x+2\right|+c\)
  3. \(\dfrac{1}{1-x}+\ln((1-x)^{2}\left|x+1\right|^{3}+c)\)

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