Integration Using Partial Fractions

You know how to add fractions: $\dfrac{2}{x - 2} + \dfrac{3}{x + 1}$ combines into one fraction, $\dfrac{5x - 4}{(x - 2)(x + 1)}$. Partial fractions runs that backwards. Integrating the combined fraction looks hard, but each simple piece just integrates to a logarithm. In AP Calculus BC, partial fractions are used for denominators that factor into distinct (non-repeating) linear factors, and they are the key step in solving the logistic differential equation.

Key ideas

The setup

If the denominator factors into different linear factors, write one simple fraction for each factor, with an unknown constant on top:

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

Three factors means three constants: $\dfrac{\ldots}{x(x - 1)(x + 2)} = \dfrac{A}{x} + \dfrac{B}{x - 1} + \dfrac{C}{x + 2}$.

This only works directly when the fraction is proper: the degree of the top is less than the degree of the bottom.

Finding the constants

Multiply both sides by the whole denominator to clear the fractions:

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

This must be true for every $x$, so substitute the values that make each factor zero:

• $x = 2$: $\;6 = 3A$, so $A = 2$.
• $x = -1$: $\;-9 = -3B$, so $B = 3$.

(You can also expand and match the coefficients of $x$ and the constants. That gives the same answer, but the zero-substitution method is usually faster.)

Integrating the pieces

Each piece is a logarithm, using $\displaystyle\int \frac{1}{x - a}\,dx = \ln|x - a| + C$:

$$\int \left(\frac{2}{x - 2} + \frac{3}{x + 1}\right)dx = 2\ln|x - 2| + 3\ln|x + 1| + C$$

Keep the absolute value bars. You can combine logs with the log laws if you like, for example $\ln|x - 2| - \ln|x + 2| = \ln\left|\dfrac{x - 2}{x + 2}\right|$.

When to divide first

If the top’s degree is greater than or equal to the bottom’s, do long division first. You get a polynomial plus a proper fraction, and only the proper fraction needs partial fractions.

Choosing a technique

Before splitting, check for something quicker. If the numerator is a constant multiple of the derivative of the denominator, $u$-substitution gives a log in one step: $\displaystyle\int \frac{2x + 1}{x^2 + x - 6}\,dx = \ln\left|x^2 + x - 6\right| + C$. Partial fractions are for when the denominator factors but substitution doesn’t fit. (A denominator that doesn’t factor, like $x^2 + 4$, calls for an arctangent instead.)

Worked examples

Example 1: Two linear factors

Find $\displaystyle\int \frac{5x - 4}{(x - 2)(x + 1)}\,dx$.

Solution. From the Key ideas, $\dfrac{5x - 4}{(x - 2)(x + 1)} = \dfrac{2}{x - 2} + \dfrac{3}{x + 1}$.

Check by recombining: $\dfrac{2(x + 1) + 3(x - 2)}{(x - 2)(x + 1)} = \dfrac{5x - 4}{(x - 2)(x + 1)}$. ✓

$$\int \frac{5x - 4}{(x - 2)(x + 1)}\,dx = 2\ln|x - 2| + 3\ln|x + 1| + C$$

Example 2: Factor first

Find $\displaystyle\int \frac{4}{x^2 - 4}\,dx$.

Solution. Factor the difference of squares: $x^2 - 4 = (x - 2)(x + 2)$. Set up

$$\frac{4}{(x - 2)(x + 2)} = \frac{A}{x - 2} + \frac{B}{x + 2} \quad\Rightarrow\quad 4 = A(x + 2) + B(x - 2)$$

• $x = 2$: $\;4 = 4A$, so $A = 1$.
• $x = -2$: $\;4 = -4B$, so $B = -1$.

$$\int \left(\frac{1}{x - 2} - \frac{1}{x + 2}\right)dx = \ln|x - 2| - \ln|x + 2| + C = \ln\left|\frac{x - 2}{x + 2}\right| + C$$

Example 3: Long division comes first

Find $\displaystyle\int \frac{x^2 + 3}{x^2 - 1}\,dx$.

Solution. Top and bottom both have degree 2, so divide first. Since $x^2 + 3 = (x^2 - 1) + 4$,

$$\frac{x^2 + 3}{x^2 - 1} = 1 + \frac{4}{x^2 - 1} = 1 + \frac{4}{(x - 1)(x + 1)}$$

Now split the proper part: $4 = A(x + 1) + B(x - 1)$.

• $x = 1$: $\;4 = 2A$, so $A = 2$.
• $x = -1$: $\;4 = -2B$, so $B = -2$.

$$\int \left(1 + \frac{2}{x - 1} - \frac{2}{x + 1}\right)dx = x + 2\ln|x - 1| - 2\ln|x + 1| + C$$

Example 4: A definite integral

Evaluate $\displaystyle\int_0^1 \frac{x + 5}{(x + 1)(x + 2)}\,dx$.

Solution. Write $x + 5 = A(x + 2) + B(x + 1)$.

• $x = -1$: $\;4 = A$.
• $x = -2$: $\;3 = -B$, so $B = -3$.

$$\begin{aligned} \int_0^1 \left(\frac{4}{x + 1} - \frac{3}{x + 2}\right)dx &= \Big[4\ln|x + 1| - 3\ln|x + 2|\Big]_0^1 \\ &= (4\ln 2 - 3\ln 3) - (4\ln 1 - 3\ln 2) \\ &= 7\ln 2 - 3\ln 3 \approx 1.556 \end{aligned}$$

Common mistakes

Skipping long division. Partial fractions only work on proper fractions. If the top’s degree is at least the bottom’s, divide first, or the constants won’t come out.

Forgetting to factor the denominator. $\dfrac{4}{x^2 - 4}$ doesn’t look like it has two pieces until you factor $x^2 - 4 = (x - 2)(x + 2)$.

Plugging in the wrong value. To find the constant over $x + 2$, substitute $x = -2$ (the value that makes $x + 2$ zero), not $x = 2$.

Losing the coefficient of x inside the log. $\displaystyle\int \frac{1}{2x + 1}\,dx = \tfrac{1}{2}\ln|2x + 1| + C$, not $\ln|2x + 1| + C$. Divide by the coefficient of $x$, as with any linear substitution.

Dropping absolute values or the + C. $\ln|x - 2|$ needs its absolute value bars, since $x - 2$ can be negative. And indefinite answers still need $+ C$.

Splitting when substitution is quicker. If the top is the derivative of the bottom (or a constant multiple of it), $u$-substitution gives the answer in one line. Both methods give equivalent answers, but one is much faster.

Practice

1. (Warm-up) Write $\dfrac{5}{(x - 1)(x + 4)}$ as a sum of partial fractions.

Solution

$5 = A(x + 4) + B(x - 1)$. At $x = 1$: $5 = 5A$, so $A = 1$. At $x = -4$: $5 = -5B$, so $B = -1$.

$$\frac{5}{(x - 1)(x + 4)} = \frac{1}{x - 1} - \frac{1}{x + 4}$$

2. (Warm-up) Find $\displaystyle\int \frac{1}{x(x + 1)}\,dx$.

Solution

$1 = A(x + 1) + Bx$. At $x = 0$: $A = 1$. At $x = -1$: $1 = -B$, so $B = -1$.

$$\int \left(\frac{1}{x} - \frac{1}{x + 1}\right)dx = \ln|x| - \ln|x + 1| + C = \ln\left|\frac{x}{x + 1}\right| + C$$

3. (Warm-up) Find $\displaystyle\int \frac{3}{x^2 - x - 2}\,dx$.

Solution

Factor: $x^2 - x - 2 = (x - 2)(x + 1)$. Then $3 = A(x + 1) + B(x - 2)$. At $x = 2$: $3 = 3A$, $A = 1$. At $x = -1$: $3 = -3B$, $B = -1$.

$$\int \left(\frac{1}{x - 2} - \frac{1}{x + 1}\right)dx = \ln|x - 2| - \ln|x + 1| + C$$

4. (Core) Find $\displaystyle\int \frac{x + 1}{x^2 + 5x + 6}\,dx$.

Solution

Factor: $x^2 + 5x + 6 = (x + 2)(x + 3)$. Then $x + 1 = A(x + 3) + B(x + 2)$.

• $x = -2$: $-1 = A$.
• $x = -3$: $-2 = -B$, so $B = 2$.

$$\int \left(-\frac{1}{x + 2} + \frac{2}{x + 3}\right)dx = -\ln|x + 2| + 2\ln|x + 3| + C$$

5. (Core) Evaluate $\displaystyle\int_3^5 \frac{2}{x^2 - 4}\,dx$.

Solution

$2 = A(x + 2) + B(x - 2)$. At $x = 2$: $A = \tfrac{1}{2}$. At $x = -2$: $B = -\tfrac{1}{2}$.

$$\begin{aligned} \int_3^5 \frac{2}{x^2 - 4}\,dx &= \frac{1}{2}\Big[\ln|x - 2| - \ln|x + 2|\Big]_3^5 \\ &= \frac{1}{2}\Big[(\ln 3 - \ln 7) - (\ln 1 - \ln 5)\Big] \\ &= \frac{1}{2}\ln\frac{15}{7} \approx 0.381 \end{aligned}$$

6. (Core) Find $\displaystyle\int \frac{x^3}{x^2 - 1}\,dx$.

Solution

The top has the larger degree, so divide: $x^3 = x(x^2 - 1) + x$, so $\dfrac{x^3}{x^2 - 1} = x + \dfrac{x}{x^2 - 1}$.

Split $\dfrac{x}{(x - 1)(x + 1)}$: $x = A(x + 1) + B(x - 1)$. At $x = 1$: $A = \tfrac{1}{2}$. At $x = -1$: $B = \tfrac{1}{2}$.

$$\int \left(x + \frac{1/2}{x - 1} + \frac{1/2}{x + 1}\right)dx = \frac{x^2}{2} + \frac{1}{2}\ln|x - 1| + \frac{1}{2}\ln|x + 1| + C$$

(The substitution $u = x^2 - 1$ also works on $\dfrac{x}{x^2 - 1}$ and gives $\tfrac{1}{2}\ln\left|x^2 - 1\right|$, which is the same thing.)

7. (Core) Find $\displaystyle\int \frac{x^2 + 2}{x(x - 1)(x + 2)}\,dx$.

Solution

Three distinct factors: $x^2 + 2 = A(x - 1)(x + 2) + Bx(x + 2) + Cx(x - 1)$.

• $x = 0$: $2 = -2A$, so $A = -1$.
• $x = 1$: $3 = 3B$, so $B = 1$.
• $x = -2$: $6 = 6C$, so $C = 1$.

$$\int \left(-\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x + 2}\right)dx = -\ln|x| + \ln|x - 1| + \ln|x + 2| + C$$

8. (Challenge) Evaluate $\displaystyle\int_0^1 \frac{x^2}{x^2 + 3x + 2}\,dx$. Give an exact answer and a decimal to 3 places.

Solution

Equal degrees, so divide: $x^2 = (x^2 + 3x + 2) - (3x + 2)$, giving $1 - \dfrac{3x + 2}{(x + 1)(x + 2)}$.

Split: $3x + 2 = A(x + 2) + B(x + 1)$. At $x = -1$: $A = -1$. At $x = -2$: $-4 = -B$, $B = 4$. So

$$\frac{x^2}{x^2 + 3x + 2} = 1 - \left(\frac{-1}{x + 1} + \frac{4}{x + 2}\right) = 1 + \frac{1}{x + 1} - \frac{4}{x + 2}$$$$\begin{aligned} \int_0^1 \left(1 + \frac{1}{x + 1} - \frac{4}{x + 2}\right)dx &= \Big[x + \ln|x + 1| - 4\ln|x + 2|\Big]_0^1 \\ &= (1 + \ln 2 - 4\ln 3) - (0 + 0 - 4\ln 2) \\ &= 1 + 5\ln 2 - 4\ln 3 \approx 0.071 \end{aligned}$$

9. (Challenge) In a logistic model you need $\displaystyle\int \frac{1}{P(10 - P)}\,dP$, where $0 \lt P \lt 10$. Find it.

Solution

$1 = A(10 - P) + BP$. At $P = 0$: $A = \tfrac{1}{10}$. At $P = 10$: $B = \tfrac{1}{10}$.

$$\int \frac{1}{P(10 - P)}\,dP = \frac{1}{10}\int \left(\frac{1}{P} + \frac{1}{10 - P}\right)dP = \frac{1}{10}\Big(\ln P - \ln(10 - P)\Big) + C = \frac{1}{10}\ln\frac{P}{10 - P} + C$$

Watch the sign: $\displaystyle\int \frac{1}{10 - P}\,dP = -\ln|10 - P|$, because the inside has derivative $-1$. Since $0 \lt P \lt 10$, both $P$ and $10 - P$ are positive, so no absolute values are needed. This is exactly the step used to solve the logistic equation.