2. Integrating Rational Functions

Video coming soon2 worked examples

Theory

To integrate a rational function, first express it in partial fractions (dividing first if it is improper), then integrate each simpler piece. Linear factors give logarithms; some quadratic denominators give an inverse tangent.

The Golden Rule: decompose into partial fractions before integrating — a term $\frac{A}{x-a}$ integrates to $A\ln|x-a|$ .

⚠️ Common Examiner Traps

Improper fractions: divide first — you cannot decompose an improper fraction directly.
Modulus in logs: the answer is $\ln|x-a|$ , with the absolute value.
Constant of integration: don't lose the $+c$ .

Worked examples

Example 1

Find $\displaystyle\int \frac{5x - 1}{(x+1)(x-2)}\,dx$ .

Step 1: In partial fractions (as found earlier), the integrand is:

\frac{5x-1}{(x+1)(x-2)} = \frac{2}{x+1} + \frac{3}{x-2}

Step 2: Integrate term by term:

\int \frac{2}{x+1} + \frac{3}{x-2}\,dx = 2\ln|x+1| + 3\ln|x-2| + c

Example 2

Find $\displaystyle\int \frac{1}{x^2 - 1}\,dx$ .

Step 1: Factorise and decompose $\frac{1}{(x-1)(x+1)}$ :

\frac{1}{x^2 - 1} = \frac{1}{2}\!\left(\frac{1}{x-1}\right) - \frac{1}{2}\!\left(\frac{1}{x+1}\right)

Step 2: Integrate each term:

= \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + c = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + c