6.12 Integrating Using Linear Partial Fractions

When to use partial fractions

Use linear partial fraction decomposition when integrating a rational function whose denominator factors into distinct linear factors, and the degree of the numerator is less than the denominator. Rewrite the fraction as a sum of simpler fractions with unknown constants (like A and B) over each linear factor, then solve for those constants before integrating.

Setting up the decomposition

For (px+q)/[(x-a)(x-b)], write it as A/(x-a) + B/(x-b). Multiply both sides by the common denominator to clear fractions, then solve for A and B using substitution (plug in x = a and x = b) or by matching coefficients of like terms.

Integrating the pieces

Once decomposed, each term integrates to a natural log: ∫A/(x-a) dx = A·ln|x-a| + C. Sum the results for all terms, combining constants of integration into one +C at the end. Always check the original degree condition first — if numerator degree ≥ denominator degree, do polynomial long division first.

Key terms

Partial fraction decomposition

Rewriting a rational function as a sum of simpler fractions, each with a linear (or other) factor as denominator.

Setup for (x)/[(x-a)(x-b)]

A/(x-a) + B/(x-b), where A and B are constants to solve for.

Integral of A/(x-a)

A·ln|x-a| + C

Condition for using linear partial fractions

Denominator factors into distinct linear factors, and numerator degree < denominator degree.

What to do if numerator degree ≥ denominator degree

Perform polynomial long division first to get a proper rational function before decomposing.

Try an AP-style question

Which is the partial fraction decomposition of 1/[(x − 1)(x + 2)]?

1. 3/(x − 1) − 3/(x + 2)
2. (1/3)/(x − 1) + (1/3)/(x + 2)
3. (1/3)/(x − 1) − (1/3)/(x + 2)
4. 1/(x − 1) − 1/(x + 2)