Linear partial fractions and logarithmic antiderivatives

Understand the idea

Partial fractions rewrites a rational function as a sum of simpler rational functions. When the denominator factors into distinct linear factors, each resulting term has a constant numerator and a logarithmic antiderivative. The decomposition is an algebraic identity on the original function's domain.

First compare degrees. If the numerator's degree is at least the denominator's, perform polynomial division before decomposing the proper rational remainder. Factor the denominator completely enough to identify the linear factors; do not cancel terms that are joined by addition.

After clearing denominators, solve for the unknown constants by matching coefficients or substituting convenient values. Integrate each fraction separately and use absolute values in logarithms. The antiderivative is defined on intervals that do not cross a zero of the original denominator.

\frac{Ax+B}{(x-a)(x-b)}=\frac{P}{x-a}+\frac{Q}{x-b},\qquad \int\frac{dx}{x-a}=\ln|x-a|+C

A method you can apply

Make the rational function proper, then factor its denominator.
Write one constant numerator over each distinct linear factor and solve the resulting identity.
Integrate the pieces and check both the derivative and the excluded input values.

Worked example

Find $\displaystyle \int \frac{3x+5}{(x+1)(x+2)}\,dx$ .

Write the integrand as $\displaystyle P/(x+1)+Q/(x+2)$ .
Clearing denominators gives $\displaystyle 3x+5=P(x+2)+Q(x+1)$ .
Matching coefficients gives $\displaystyle P+Q=3$ and $\displaystyle 2P+Q=5$ , so $\displaystyle P=2$ and $\displaystyle Q=1$ .
Integrate $\displaystyle 2/(x+1)+1/(x+2)$ term by term.

The result is

\displaystyle 2\ln|x+1|+\ln|x+2|+C

, on an interval excluding

\displaystyle -1

and

\displaystyle -2

.

Try it yourself

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

Show solution and reasoning

Use $\displaystyle P/(x-1)+Q/(x+2)$ . The equations are $\displaystyle P+Q=5$ and $\displaystyle 2P-Q=1$ , giving $\displaystyle P=2$ and $\displaystyle Q=3$ . Therefore the antiderivative is $\displaystyle 2\ln|x-1|+3\ln|x+2|+C$ .