Integration by Partial Fractions (Part 1)

This lesson covers integration of rational functions using partial fractions. We work through three examples: distinct linear factors, a repeated factor combined with a distinct factor, and a case where the denominator must be factorised first.

Example 1: distinct linear factors

We want to evaluate

$I = \int \frac{1}{x^2 - 9}\,dx = \int \frac{1}{(x+3)(x-3)}\,dx.$

Write the integrand as a sum of partial fractions:

$\frac{1}{(x+3)(x-3)} = \frac{A}{x+3} + \frac{B}{x-3}.$

Taking the common denominator and cancelling it gives

$1 = A(x-3) + B(x+3).$

Find $B$. Put $x = 3$: $\;1 = B(6)$, so $B = \tfrac{1}{6}$.

Find $A$. Put $x = -3$: $\;1 = A(-6)$, so $A = -\tfrac{1}{6}$.

Therefore

$I = -\tfrac{1}{6}\int \frac{dx}{x+3} + \tfrac{1}{6}\int \frac{dx}{x-3} = -\tfrac{1}{6}\ln|x+3| + \tfrac{1}{6}\ln|x-3| + C.$

Combining the logarithms,

$I = \tfrac{1}{6}\ln\left|\frac{x-3}{x+3}\right| + C.$

Example 2: a repeated factor with a distinct factor

Next we evaluate

$I = \int \frac{x}{(x-1)^2(x+2)}\,dx.$

A repeated factor needs two terms, so we set up

$\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}.$

Clearing denominators gives

$x = A(x-1)(x+2) + B(x+2) + C(x-1)^2.$

Find $B$. Put $x = 1$: $\;1 = B(3)$, so $B = \tfrac{1}{3}$.

Find $C$. Put $x = -2$: $\;-2 = C(-3)^2 = 9C$, so $C = -\tfrac{2}{9}$.

Find $A$. Compare the coefficients of $x^2$. On the right, $x^2$ comes from $A(x-1)(x+2)$ and $C(x-1)^2$, giving $A + C = 0$. Hence $A = -C = \tfrac{2}{9}$.

So

$\frac{x}{(x-1)^2(x+2)} = \frac{2/9}{x-1} + \frac{1/3}{(x-1)^2} - \frac{2/9}{x+2}.$

Integrating term by term, and using $\int \frac{dx}{(x-1)^2} = -\frac{1}{x-1}$,

$I = \tfrac{2}{9}\ln|x-1| - \frac{1}{3(x-1)} - \tfrac{2}{9}\ln|x+2| + C.$

Combining the logarithms,

$I = \tfrac{2}{9}\ln\left|\frac{x-1}{x+2}\right| - \frac{1}{3(x-1)} + C.$

Example 3: factorising the cubic denominator first

Finally we evaluate

$I = \int \frac{3x+5}{x^3 - x^2 - x + 1}\,dx.$

Factorise the denominator by grouping:

$x^3 - x^2 - x + 1 = x^2(x-1) - (x-1) = (x-1)(x^2-1) = (x-1)^2(x+1).$

So we need

$\frac{3x+5}{(x+1)(x-1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{(x-1)^2}.$

Clearing denominators,

$3x+5 = A(x-1)^2 + B(x-1)(x+1) + C(x+1).$

Find $C$. Put $x = 1$: $\;8 = C(2)$, so $C = 4$.

Find $A$. Put $x = -1$: $\;2 = A(-2)^2 = 4A$, so $A = \tfrac{1}{2}$.

Find $B$. Compare constant terms: $5 = A - B + C = \tfrac{1}{2} - B + 4$, so $B = -\tfrac{1}{2}$.

Therefore

$\frac{3x+5}{(x+1)(x-1)^2} = \frac{1/2}{x+1} - \frac{1/2}{x-1} + \frac{4}{(x-1)^2}.$

Integrating,

$I = \tfrac{1}{2}\ln|x+1| - \tfrac{1}{2}\ln|x-1| - \frac{4}{x-1} + C,$

which combines to

$I = \tfrac{1}{2}\ln\left|\frac{x+1}{x-1}\right| - \frac{4}{x-1} + C.$

Key takeaways

• Split a rational function into partial fractions, one term per factor, with a repeated factor $(x-a)^2$ contributing both a $\frac{1}{x-a}$ and a $\frac{1}{(x-a)^2}$ term.
• Find the constants quickly by substituting values that zero out factors, then compare coefficients for any remaining unknowns.
• Use $\int \frac{dx}{x-a} = \ln|x-a|$ and $\int \frac{dx}{(x-a)^2} = -\frac{1}{x-a}$, and combine logarithms at the end.