Math · Calculus II · Worked example

Partial fractions with a repeated factor

Find ∫(2x² + 3)/(x(x + 1)²) dx.

Set up the terms

The factor x gets A/x, and the repeated factor (x + 1)² gets B/(x + 1) + C/(x + 1)². Clearing denominators gives 2x² + 3 = A(x + 1)² + Bx(x + 1) + Cx.

Substitute convenient values

x = 0 gives 3 = A, and x = −1 gives 5 = −C, so C = −5.

\begin{array}{l} \frac{2 {(0)}^{2} + 3}{{(0 + 1)}^{2}} = 3 \\ \frac{2 {(- 1)}^{2} + 3}{- 1} = - 5 \end{array}

Match a coefficient

The x² terms give 2 = A + B, so B = 2 − 3 = −1.

2 - 3 = - 1

Check and integrate

Recombining returns the original fraction. The last term integrates as a power: ∫−5/(x + 1)² dx = 5/(x + 1).

\begin{array}{l} \frac{2 x^{2} + 3}{x {(x + 1)}^{2}} \\ \frac{3}{x} - \frac{1}{x + 1} - \frac{5}{{(x + 1)}^{2}} \end{array}

Result

3 ln|x| − ln|x + 1| + 5/(x + 1) + C.

Your turn

Decompose (x + 3)/(x − 1)².

Show the answer and explanation

1/(x − 1) + 4/(x − 1)².

x + 3 = A(x − 1) + B. Setting x = 1 gives B = 4, and the x terms give A = 1.

\begin{array}{l} \frac{x + 3}{{(x - 1)}^{2}} \\ \frac{1}{x - 1} + \frac{4}{{(x - 1)}^{2}} \end{array}

Keep exploring

Derivative & antiderivative checks verifies the antiderivative on x > 0.

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

OpenStax Calculus Volume 2 — Partial fractions

Make it concrete

Try in the workspace

Open the example inputs, change a value and keep a useful result on your board.

Check the antiderivative Check the split in Math Open worked example on a board Integral of 1/x in Math Reference

Your existing work stays on this device. Examples open as editable copies.