Integration by partial fractions

A rational function $\dfrac{P(x)}{Q(x)}$ can be integrated, but the techniques depend on its shape. Partial fractions is the algorithm for transforming the integrand into a sum of fractions whose denominators are linear or irreducible quadratic , each of which has a known antiderivative.

When does partial fractions apply?

The integrand must be a proper rational function: $\deg P < \deg Q$. If the integrand is improper, perform polynomial division first to write $\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}, \qquad \deg R < \deg Q.$ Then $\int S$ is a polynomial integral (trivial), and we apply partial fractions to $\dfrac{R}{Q}$.

The four cases

After factoring $Q(x)$ over the reals into linear and irreducible-quadratic factors, the decomposition depends on the shapes of the factors.

Case 1: Distinct linear factors. If $Q(x) = (x - a_1)(x - a_2)\cdots(x - a_n)$ with distinct $a_i$, then $\frac{P(x)}{Q(x)} = \frac{A_1}{x - a_1} + \frac{A_2}{x - a_2} + \cdots + \frac{A_n}{x - a_n}.$

Case 2: Repeated linear factor. If $(x - a)^k$ divides $Q$, the decomposition includes $\frac{A_1}{x - a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_k}{(x - a)^k}.$

Case 3: Distinct irreducible quadratic factor. For each $(x^2 + bx + c)$ factor of $Q$ with $b^2 - 4c < 0$, include $\frac{Bx + C}{x^2 + bx + c}.$

Case 4: Repeated irreducible quadratic. For $(x^2 + bx + c)^k$, include $\frac{B_1 x + C_1}{x^2 + bx + c} + \frac{B_2 x + C_2}{(x^2 + bx + c)^2} + \cdots + \frac{B_k x + C_k}{(x^2 + bx + c)^k}.$

Finding the unknowns

Two methods. Method 1 , clear denominators and compare coefficients. Multiply both sides by $Q(x)$, expand, and equate coefficients of like powers of $x$. This always works but can be slow.

Method 2 , Heaviside cover-up. For a distinct linear factor $x - a$, the coefficient $A$ satisfies $A = \lim_{x \to a} (x - a) \cdot \frac{P(x)}{Q(x)} = \frac{P(a)}{Q'(a)} \;\text{(when convenient)}.$ In practice, "cover up" the $(x - a)$ in the denominator and substitute $x = a$ into what remains.

For repeated factors and quadratic factors, you typically need Method 1 or a hybrid.

Integrating each piece

Once decomposed, each piece is one of:

• $\int \dfrac{A}{x - a}\,dx = A\ln|x - a| + C$
• $\int \dfrac{A}{(x-a)^k}\,dx = -\dfrac{A}{(k-1)(x-a)^{k-1}} + C$ for $k \ge 2$
• $\int \dfrac{Bx + C}{x^2 + bx + c}\,dx$: split into $\dfrac{B}{2}\cdot\dfrac{2x+b}{x^2+bx+c}$ (a log) plus $\dfrac{C - Bb/2}{x^2+bx+c}$ (complete the square, $\tan^{-1}$).

Worked examples

Example 1. $\int \dfrac{1}{(x-1)(x-2)}\,dx$.

Decompose: $\dfrac{1}{(x-1)(x-2)} = \dfrac{A}{x-1} + \dfrac{B}{x-2}$. Cover-up: $A = \dfrac{1}{1-2} = -1$, $B = \dfrac{1}{2-1} = 1$. So $\int \dfrac{dx}{(x-1)(x-2)} = -\ln|x-1| + \ln|x-2| + C = \ln\left|\dfrac{x-2}{x-1}\right| + C.$

Example 2. $\int \dfrac{x+1}{x^2 - 4}\,dx$.

$x^2 - 4 = (x-2)(x+2)$. Decompose: $\dfrac{x+1}{(x-2)(x+2)} = \dfrac{A}{x-2} + \dfrac{B}{x+2}$. Cover-up: $A = \dfrac{3}{4}$, $B = \dfrac{-1}{-4} = \dfrac{1}{4}$. Integral $= \dfrac{3}{4}\ln|x-2| + \dfrac{1}{4}\ln|x+2| + C$.

Example 3. $\int \dfrac{1}{x(x-1)^2}\,dx$.

Decompose: $\dfrac{1}{x(x-1)^2} = \dfrac{A}{x} + \dfrac{B}{x-1} + \dfrac{C}{(x-1)^2}$. Multiply by $x(x-1)^2$: $1 = A(x-1)^2 + Bx(x-1) + Cx$. Set $x = 0$: $1 = A$. Set $x = 1$: $1 = C$. Compare coefficients of $x^2$: $0 = A + B$, so $B = -1$. Integral $= \ln|x| - \ln|x-1| - \dfrac{1}{x-1} + C$.

Example 4. $\int \dfrac{x^2}{(x-1)(x+1)^2}\,dx$.

Decompose: $\dfrac{x^2}{(x-1)(x+1)^2} = \dfrac{A}{x-1} + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2}$. Multiply through: $x^2 = A(x+1)^2 + B(x-1)(x+1) + C(x-1)$. Set $x = 1$: $1 = 4A$, so $A = 1/4$. Set $x = -1$: $1 = -2C$, so $C = -1/2$. Compare coefficients of $x^2$: $1 = A + B$, so $B = 3/4$. Integral $= \dfrac{1}{4}\ln|x-1| + \dfrac{3}{4}\ln|x+1| + \dfrac{1}{2(x+1)} + C$.

Example 5. $\int \dfrac{x}{(x^2+1)(x-1)}\,dx$.

Decompose: $\dfrac{x}{(x^2+1)(x-1)} = \dfrac{Ax+B}{x^2+1} + \dfrac{C}{x-1}$. Multiply: $x = (Ax+B)(x-1) + C(x^2+1)$. Set $x = 1$: $1 = 2C$, so $C = 1/2$. Expand: $x = Ax^2 - Ax + Bx - B + Cx^2 + C$. Coefficient of $x^2$: $0 = A + C$, so $A = -1/2$. Coefficient of $x^0$: $0 = -B + C$, so $B = 1/2$. Integral splits as $-\dfrac{1}{4}\ln(x^2+1) + \dfrac{1}{2}\tan^{-1}x + \dfrac{1}{2}\ln|x-1| + C$.

Example 6. $\int \dfrac{x^3 + 1}{x^2 - 1}\,dx$.

Improper , divide first: $\dfrac{x^3+1}{x^2-1} = x + \dfrac{x+1}{x^2-1} = x + \dfrac{1}{x-1}$ (since $\dfrac{x+1}{(x-1)(x+1)} = \dfrac{1}{x-1}$). Integral $= \dfrac{x^2}{2} + \ln|x-1| + C$.

Try it yourself

1. $\int \dfrac{1}{x^2 - 9}\,dx$
2. $\int \dfrac{5}{(x-1)(x+4)}\,dx$
3. $\int \dfrac{2x+3}{x^2 + 3x + 2}\,dx$
4. $\int \dfrac{1}{x^2 - 5x + 6}\,dx$
5. $\int \dfrac{x^2+1}{(x+1)^2(x-2)}\,dx$
6. $\int \dfrac{1}{x(x^2+1)}\,dx$
7. $\int \dfrac{3x+1}{(x-1)(x^2+1)}\,dx$
8. $\int \dfrac{x^2}{x^2 - 1}\,dx$ (improper)
9. $\int \dfrac{2x^3 + 1}{x^2 + x}\,dx$
10. $\int \dfrac{x}{(x-1)^3}\,dx$
11. $\int \dfrac{dx}{x(x^4 + 1)}$ (substitute $u = x^4$ and then partial fractions)
12. $\int \dfrac{dx}{\sin x(1 - \cos x)}$ (Weierstrass substitution $t = \tan(x/2)$)
13. $\int \dfrac{e^x}{(e^x - 1)(e^x + 2)}\,dx$ (substitute $u = e^x$)
14. $\int \dfrac{x^4}{(x-1)(x^2+1)}\,dx$

Pitfalls and tricks

• Always check degree before decomposing. Improper fractions need division first.
• Factor the denominator completely over the reals. Linear or irreducible-quadratic , those are the only legal pieces.
• For repeated factors, include every power up to the maximum, not just the highest.
• Heaviside cover-up is fast for distinct linear factors. Use it whenever you can.
• Watch the signs when integrating each piece , a misplaced minus is the most common error.