Integration by Partial Fractions

The idea

When the denominator of a rational function factors into distinct linear pieces, the whole fraction can be rewritten as a sum of simpler fractions, each of which has a known antiderivative involving a logarithm:

Partial-fraction forms

The form to write depends on the factors of the denominator. This only works when the fraction is proper (the numerator's degree is lower than the denominator's). If it isn't, divide first.

Worked example

Find $\displaystyle\int \dfrac{x}{(x+1)(x+2)}\,dx$.

Solution: Write $\dfrac{x}{(x+1)(x+2)} = \dfrac{A}{x+1} + \dfrac{B}{x+2}$, so $x = A(x+2) + B(x+1)$. Setting $x=-1$ gives $A=-1$; setting $x=-2$ gives $B=2$:

Worked example

Find $\displaystyle\int \dfrac{2x}{x^2+3x+2}\,dx$.

Solution: The denominator factors as $(x+1)(x+2)$. Writing $\dfrac{2x}{(x+1)(x+2)} = \dfrac{A}{x+1} + \dfrac{B}{x+2}$ gives $2x = A(x+2)+B(x+1)$. Setting $x=-1$: $A=-2$. Setting $x=-2$: $B=4$:

Worked example: a repeated factor

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

Solution: Write $\dfrac{1}{(x-1)^2(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{D}{x+2}$, so $1 = A(x-1)(x+2) + B(x+2) + D(x-1)^2$. Setting $x = 1$ gives $B = \dfrac{1}{3}$, and setting $x = -2$ gives $D = \dfrac{1}{9}$. Comparing the $x^2$ coefficients, $0 = A + D$, so $A = -\dfrac{1}{9}$:

The repeated-factor term integrates as a power, not a logarithm: $\int (x-1)^{-2}\,dx = -(x-1)^{-1}$.

Worked example: a quadratic factor

Find $\displaystyle\int \dfrac{2x}{(x+1)(x^2+1)}\,dx$.

Solution: $x^2 + 1$ has no real factors, so its numerator is linear: $\dfrac{2x}{(x+1)(x^2+1)} = \dfrac{A}{x+1} + \dfrac{Bx + D}{x^2+1}$. Then $2x = A(x^2+1) + (Bx + D)(x+1)$. Setting $x = -1$ gives $A = -1$. Comparing $x^2$ coefficients, $0 = A + B$, so $B = 1$. Comparing constants, $0 = A + D$, so $D = 1$:

Common mistakes

• Skipping the division step. $\dfrac{x^2}{(x+1)(x+2)}$ is improper (degree 2 over degree 2), so divide first: it equals $1 - \dfrac{3x + 2}{(x+1)(x+2)}$.
• Writing only one term for a repeated factor. $(x-1)^2$ needs both $\dfrac{A}{x-1}$ and $\dfrac{B}{(x-1)^2}$.
• Putting a constant over an irreducible quadratic. Over $x^2 + 1$ the numerator must be $Bx + C$, not just $B$.