Partial fractions in integration: split, then log

8 minA Level3.5.3

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can integrate a fraction like $\dfrac{3}{(x-1)(2x+1)}$ by splitting it into partial fractions (from topic 3.1) and integrating each piece.

Why split it up?

$\displaystyle\int \frac{3}{(x-1)(2x+1)}\,dx$ isn't a standard integral. You can't integrate the top and bottom separately.

But in topic 3.1 you learned to split a fraction like this into simpler ones:

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

Each simple piece is a standard integral. So:

Split into partial fractions.
Integrate each piece.
Tidy up, often into a single log, then put in any limits.

Definition

The integrals of the pieces

$k$ , $a$ and $b$ are constants, with $a \neq 0$ .

A linear bottom gives a log. Divide by $a$ , the number in front of $x$ :

$\int \frac{k}{ax + b}\,dx = \frac{k}{a}\ln|ax + b| + C$

A squared linear bottom gives a fraction, using the power rule on $(ax + b)^{-2}$ :

$\int \frac{k}{(ax + b)^2}\,dx = -\frac{k}{a(ax + b)} + C$

A bottom of $x^2 + c^2$ splits into a log part and an inverse-tan part:

$\int \frac{x}{x^2 + c^2}\,dx = \tfrac12\ln(x^2 + c^2) + C, \qquad \int \frac{1}{x^2 + c^2}\,dx = \frac{1}{c}\tan^{-1}\frac{x}{c} + C$

The bars $|\ \ |$ mean "the size of", so the log works even when the inside is negative.

Worked example: Example 1: two different linear factors

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

Have a go first?

Quick check

Multiple choiceQuestion 1 · Recall

Answer the 3 questions above to continue.