Partial fractions with a quadratic factor

When the denominator has a linear factor and a quadratic factor \(x^2+c^2\) that cannot be factorised, write \(\dfrac{px^2+qx+r}{(ax+b)(x^2+c^2)}=\dfrac{A}{ax+b}+\dfrac{Bx+C}{x^2+c^2}\).

Why the numerator is \(Bx+C\): a proper fraction over a quadratic can have a numerator of degree one, so it needs both an \(x\) term and a constant. \(x^2+c^2\) has no real roots, so it cannot be split further. A quadratic such as \(2x^2+1\) or \(3x^2+1\) is of the same type: write \(\dfrac{Bx+C}{2x^2+1}\).

Method

1. Write the form with \(Bx+C\) over the quadratic factor.
2. Multiply through by the whole denominator.
3. Substitute the root of the linear factor to find \(A\).
4. Compare the coefficients of \(x^2\) to find \(B\) and the constant terms to find \(C\); check with the coefficient of \(x\).

Example

Express \(\dfrac{x^2+5x+4}{(2x-1)(x^2+2)}\) in partial fractions.

\(\dfrac{x^2+5x+4}{(2x-1)(x^2+2)}=\dfrac{A}{2x-1}+\dfrac{Bx+C}{x^2+2}\), so \(x^2+5x+4=A(x^2+2)+(Bx+C)(2x-1)\).

\(x=\dfrac12\): \(\dfrac{27}{4}=\dfrac94A\), \(A=3\). \(x^2\): \(1=A+2B\), \(B=-1\). Constants: \(4=2A-C\), \(C=2\). Check \(x\): \(-B+2C=5\).

\(\therefore\ \dfrac{x^2+5x+4}{(2x-1)(x^2+2)}=\dfrac{3}{2x-1}+\dfrac{2-x}{x^2+2}\).