Partial fractions with a repeated linear factor

When a linear factor is repeated in the denominator, write one fraction for each power of it: \(\dfrac{px^2+qx+r}{(ax+b)(cx+d)^2}=\dfrac{A}{ax+b}+\dfrac{B}{cx+d}+\dfrac{C}{(cx+d)^2}\).

Why two terms for \((cx+d)^2\): together \(\dfrac{B}{cx+d}+\dfrac{C}{(cx+d)^2}=\dfrac{B(cx+d)+C}{(cx+d)^2}\), a linear numerator over \((cx+d)^2\). With \(\dfrac{C}{(cx+d)^2}\) alone the numerator would be a constant, which cannot match a general numerator. A cubic denominator needs three constants.

Method

1. Factorise the denominator fully and write the form with both powers of the repeated factor. A factor \(x^2\) is the repeated factor \(x\): use \(\dfrac{B}{x}+\dfrac{C}{x^2}\).
2. Multiply through by the whole denominator.
3. Substitute the root of each linear factor to find \(A\) and \(C\).
4. Compare the coefficients of \(x^2\) (or substitute \(x=0\)) to find \(B\).

Example

Express \(\dfrac{5x+4}{x(x+2)^2}\) in partial fractions.

\(\dfrac{5x+4}{x(x+2)^2}=\dfrac{A}{x}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}\), so \(5x+4=A(x+2)^2+Bx(x+2)+Cx\).

\(x=0\): \(4=4A\), \(A=1\). \(x=-2\): \(-6=-2C\), \(C=3\). Coefficient of \(x^2\): \(0=A+B\), \(B=-1\).

\(\therefore\ \dfrac{5x+4}{x(x+2)^2}=\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{3}{(x+2)^2}\). Check at \(x=1\): \(\dfrac99=1\) and \(1-\dfrac13+\dfrac39=1\).