Linear Partial Fractions

Adding fractions combines them over a common denominator. Partial fractions runs that in reverse: a fraction with a factored denominator is split back into simpler fractions, each of which integrates to a logarithm.

The decomposition

To find the constants, multiply both sides by

Q(x)

. The result,

P(x) = A(a_2x + b_2) + B(a_1x + b_1)

, is called the basic equation. Both sides are polynomials that agree everywhere except possibly at the two zeros, so they agree there too. Substituting the zero of

a_1x + b_1

makes the

B

term 0 and gives

A

, and the other zero gives

B

Example 1

Find each indefinite integral.

\begin{aligned} &\text{(a)}\;\; \int \frac{5x - 4}{x^2 - x - 2}\,dx \\[10pt] &\text{(b)}\;\; \int \frac{3x + 2}{2x^2 + 5x - 3}\,dx \end{aligned}

(a) Factor:

x^2 - x - 2 = (x - 2)(x + 1)

, so the form is

\frac{5x - 4}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1}

. The basic equation is

5x - 4 = A(x + 1) + B(x - 2).

x = 2

6 = 3A

, so

A = 2

. At

x = -1

-9 = -3B

, so

B = 3

\begin{aligned} &\int \frac{5x - 4}{x^2 - x - 2}\,dx \\[4pt] &\quad = \int \left(\frac{2}{x - 2} + \frac{3}{x + 1}\right)dx \\[4pt] &\quad = 2\ln|x - 2| + 3\ln|x + 1| + C. \end{aligned}

(b) Factor:

2x^2 + 5x - 3 = (2x - 1)(x + 3)

, so the form is

\frac{A}{2x - 1} + \frac{B}{x + 3}

. The basic equation is

3x + 2 = A(x + 3) + B(2x - 1).

The zero of

2x - 1

x = \frac{1}{2}

, where

\frac{7}{2} = \frac{7}{2}A

, so

A = 1

. At

x = -3

-7 = -7B

, so

B = 1

. The factor

2x - 1

needs the substitution

u = 2x - 1

, with

du = 2\,dx

, which brings out

\frac{1}{2}

\begin{aligned} &\int \frac{3x + 2}{2x^2 + 5x - 3}\,dx \\[4pt] &\quad = \int \left(\frac{1}{2x - 1} + \frac{1}{x + 3}\right)dx \\[4pt] &\quad = \frac{1}{2}\ln|2x - 1| + \ln|x + 3| + C. \end{aligned}

A factor written with

x

last, such as

4 - x

, has derivative

-1

. Its integral is

\int \frac{dx}{4 - x} = -\ln|4 - x| + C

Example 2

Evaluate the integral below.

\int_1^2 \frac{6}{(x + 2)(4 - x)}\,dx

The form is

\frac{A}{x + 2} + \frac{B}{4 - x}

, and the basic equation is

6 = A(4 - x) + B(x + 2)

. At

x = 4

6 = 6B

, so

B = 1

. At

x = -2

6 = 6A

, so

A = 1

. Both factors are positive on

[1, 2]

\begin{aligned} &\int_1^2 \left(\frac{1}{x + 2} + \frac{1}{4 - x}\right)dx \\[4pt] &\qquad = \Big[\ln(x + 2) - \ln(4 - x)\Big]_1^2 \\[4pt] &\qquad = (\ln 4 - \ln 2) - (\ln 3 - \ln 3) \\[4pt] &\qquad = \ln 2 \approx 0.693 \end{aligned}

Before decomposing

Two checks come first. If the numerator is a constant multiple of the derivative of the denominator, a substitution finishes the integral at once. And if the numerator's degree is at least the denominator's, divide first and decompose only the remainder over the divisor.

Example 3

Find each indefinite integral.

\begin{aligned} &\text{(a)}\;\; \int \frac{8x - 6}{2x^2 - 3x - 5}\,dx \\[10pt] &\text{(b)}\;\; \int \frac{x^3 + 2}{x^2 - x}\,dx \end{aligned}

(a) The derivative of

2x^2 - 3x - 5

4x - 3

, and the numerator is

2(4x - 3)

. With

u = 2x^2 - 3x - 5

du = (4x - 3)\,dx

, and

\begin{aligned} &\int \frac{8x - 6}{2x^2 - 3x - 5}\,dx \\[4pt] &\qquad = 2\ln\left|2x^2 - 3x - 5\right| + C. \end{aligned}

Partial fractions gives the same answer, as

2\ln|2x - 5| + 2\ln|x + 1| + C

, with more work.

(b) The numerator has degree 3 and the denominator degree 2, so divide:

x^3 + 2 = (x^2 - x)(x + 1) + (x + 2)

. Then decompose the remainder over the divisor,

\frac{x + 2}{x(x - 1)} = \frac{A}{x} + \frac{B}{x - 1}

. The basic equation is

x + 2 = A(x - 1) + Bx

. At

x = 0

2 = -A

, so

A = -2

. At

x = 1

B = 3

\begin{aligned} &\int \frac{x^3 + 2}{x^2 - x}\,dx \\[4pt] &\quad = \int \left(x + 1 - \frac{2}{x} + \frac{3}{x - 1}\right)dx \\[4pt] &\quad = \frac{x^2}{2} + x - 2\ln|x| \\[4pt] &\qquad\quad + 3\ln|x - 1| + C. \end{aligned}