Math · Calculus II · Worked example

Integrate using partial fractions

Find ∫(5x − 1)/(x² − x − 2) dx.

Factor the denominator

x² − x − 2 = (x − 2)(x + 1): two distinct linear factors, and the fraction is proper.

Write (5x − 1)/((x − 2)(x + 1)) = A/(x − 2) + B/(x + 1). Multiplying by the denominator gives 5x − 1 = A(x + 1) + B(x − 2).

x = 2 gives 9 = 3A, and x = −1 gives −6 = −3B. Covering up each factor and substituting its zero gives the same numbers.

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

Recombining the two fractions returns the original one.

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

Each term integrates to a logarithm: 3 ln|x − 2| + 2 ln|x + 1| + C.

∫(5x − 1)/(x² − x − 2) dx = 3 ln|x − 2| + 2 ln|x + 1| + C.

Find ∫(x + 7)/(x² − x − 6) dx.

Show the answer and explanation

x² − x − 6 = (x − 3)(x + 2). At x = 3, 10 = 5A, so A = 2; at x = −2, 5 = −5B, so B = −1.

\begin{array}{l} \frac{x + 7}{x^{2} - x - 6} \\ \frac{2}{x - 3} - \frac{1}{x + 2} \end{array}

Derivative & antiderivative checks differentiates 3 ln|x − 2| + 2 ln|x + 1| on x > 2 and confirms that it returns the integrand.

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