Partial fractions

What you will be able to do

• split a fraction of polynomials into partial fractions of the right form;
• find the constants with the cover-up rule or by comparing coefficients;
• integrate each piece, dividing first when the fraction is improper.

One fraction from two

You know how to add $\dfrac{2}{x - 1} + \dfrac{1}{x + 3}$. The sum is $\dfrac{2(x + 3) + (x - 1)}{(x - 1)(x + 3)} = \dfrac{3x + 5}{(x - 1)(x + 3)}$. Partial fractions run this backwards. Starting from $\dfrac{3x + 5}{(x - 1)(x + 3)}$, we split it into the two simple parts, and each part is easy to integrate.

The forms

Let $\dfrac{P(x)}{Q(x)}$ be a fraction in which the degree of $P$ is less than the degree of $Q$. Factorise $Q$. Then each factor of $Q$ gives terms of a fixed shape.

• A factor $(x - a)$ gives the term $\dfrac{A}{x - a}$.
• A repeated factor $(x - a)^2$ gives two terms, $\dfrac{A}{x - a} + \dfrac{B}{(x - a)^2}$.
• A quadratic factor $x^2 + p^2$ gives $\dfrac{Bx + C}{x^2 + p^2}$.

A repeated factor needs one term for each power up to the repeat. A quadratic that does not factorise over the real numbers needs a top of the form $Bx + C$.

Finding the constants

The cover-up rule works for distinct linear factors. To find the constant over $(x - a)$, hide that factor in the original fraction, and put $x = a$ in what is left.

Comparing coefficients works every time. Multiply both sides by the bottom, expand, and compare the coefficients of each power of $x$. You can also put in convenient values of $x$.

The integrals

Each type of term has a known integral.

• $\displaystyle\int \dfrac{A}{x - a}\,dx = A\ln|x - a|$.
• $\displaystyle\int \dfrac{B}{(x - a)^2}\,dx = -\dfrac{B}{x - a}$.
• $\displaystyle\int \dfrac{Bx + C}{x^2 + p^2}\,dx$ splits into $\dfrac{B}{2}\ln(x^2 + p^2)$ and $\dfrac{C}{p}\tan^{-1}\dfrac{x}{p}$.

Worked examples

Example 1 (distinct factors). Find $\displaystyle\int \dfrac{3x + 5}{(x - 1)(x + 3)}\,dx$.

1. Write $\dfrac{3x + 5}{(x - 1)(x + 3)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 3}$.
2. Cover up $x - 1$ and put $x = 1$: $A = \dfrac{3 + 5}{1 + 3} = 2$.
3. Cover up $x + 3$ and put $x = -3$: $B = \dfrac{-9 + 5}{-3 - 1} = 1$.
4. The integral is $2\ln|x - 1| + \ln|x + 3| + C$.

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

1. Write $\dfrac{x^2 + 1}{(x - 1)^2(x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{(x - 1)^2} + \dfrac{C}{x + 2}$.
2. Cover up $(x - 1)^2$ and put $x = 1$: $B = \dfrac{2}{3}$. Cover up $x + 2$ and put $x = -2$: $C = \dfrac{5}{9}$.
3. For $A$, put $x = 0$ in the original: $\dfrac{1}{2} = -A + B + \dfrac{C}{2}$, so $A = \dfrac{2}{3} + \dfrac{5}{18} - \dfrac{1}{2} = \dfrac{4}{9}$.
4. The integral is $\dfrac{4}{9}\ln|x - 1| - \dfrac{2}{3(x - 1)} + \dfrac{5}{9}\ln|x + 2| + C$.

Example 3 (a quadratic factor). Find $\displaystyle\int \dfrac{dx}{(x + 1)(x^2 + 1)}$.

1. Write $\dfrac{1}{(x + 1)(x^2 + 1)} = \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 + 1}$. Cover up $x + 1$: $A = \dfrac{1}{2}$.
2. Multiply out: $1 = A(x^2 + 1) + (Bx + C)(x + 1)$. The $x^2$ terms give $0 = A + B$, so $B = -\dfrac{1}{2}$. At $x = 0$, $1 = A + C$, so $C = \dfrac{1}{2}$.
3. The integral is $\dfrac{1}{2}\ln|x + 1| - \dfrac{1}{4}\ln(x^2 + 1) + \dfrac{1}{2}\tan^{-1}x + C$.

Example 4 (divide first). Find $\displaystyle\int \dfrac{x^2}{x^2 - 1}\,dx$. The top and the bottom have the same degree, so divide.

1. $\dfrac{x^2}{x^2 - 1} = 1 + \dfrac{1}{x^2 - 1} = 1 + \dfrac{1}{2}\left(\dfrac{1}{x - 1} - \dfrac{1}{x + 1}\right)$.
2. The integral is $x + \dfrac{1}{2}\ln\left|\dfrac{x - 1}{x + 1}\right| + C$.

Practise with new numbers

Try it in Python

The program finds $A$ and $B$ for $\dfrac{3x + 5}{(x - 1)(x + 3)}$ by the cover-up rule. Then it compares the original fraction with the sum of the parts at four values of $x$.

Common mistakes

• Not dividing first. If the top has degree at least as large as the bottom, the split is wrong until you divide.
• A missing term for a repeated factor. $(x - 1)^2$ needs both $\dfrac{A}{x - 1}$ and $\dfrac{B}{(x - 1)^2}$.
• A constant top over a quadratic. An irreducible quadratic needs $Bx + C$ on top.
• Wrong signs in the cover-up rule. Put the root in every other factor, with its sign.
• Dropping the logarithm's modulus. $\ln|x - a|$ is safe for any $x$ with $x \ne a$.

Remember these

• Factorise the bottom. Each linear factor gives $\dfrac{A}{x - a}$, a repeated factor gives all its powers, and a quadratic factor gives $\dfrac{Bx + C}{x^2 + p^2}$.
• The cover-up rule finds the constants for distinct linear factors. Comparing coefficients works for all.
• If the degree of the top is at least the degree of the bottom, divide first.
• The integrals are $A\ln|x - a|$, $-\dfrac{B}{x - a}$, and a logarithm with an inverse tangent for the quadratic.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Split $\dfrac{7x - 5}{(x - 1)(x - 2)}$ into partial fractions.

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

3. What is the form of the split of $\dfrac{1}{(x - 1)(x^2 + 4)}$?

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

5. Why can $\dfrac{x^2}{(x - 1)(x + 1)}$ not be written at once as $\dfrac{A}{x - 1} + \dfrac{B}{x + 1}$?

References

• OpenStax, Calculus Volume 2, Chapter 3, section 3.4 Partial Fractions.
• NCERT, Mathematics Part II for Class XII, Chapter 7, the section on integration by partial fractions.