Blog · Calculus guides · 25 October 2026 · 4 min read

Partial Fractions: Steps, Examples & Practice Problems

Q: What if the denominator does not factor?

A quadratic like x^2+4 that will not factor gets a term (Ax+B)/(x^2+4), which integrates to a mix of ln(x^2+4) and arctanx/2. The arctan part is the territory of trig substitution.

Q: Is there a shortcut for finding the constants?

Yes, the cover-up method. For a distinct factor x-a, cover it up in the original fraction and substitute x = a into what is left. For (5x+1)/((x-1)(x+2)), covering (x-1) and putting x = 1 gives 6/3 = 2, which is A. It does not work for the lower terms of repeated factors.

Q: Why are there absolute values in the answer?

Because ln only takes positive inputs, and ∫1/u du = ln|u| + C is valid on both sides of u = 0. On an interval where x - 1 > 0 you can drop them.

$An illustration of Partial fractions$

Partial fractions integrates a fraction of polynomials by splitting it into simpler fractions you already know how to integrate. Factor the denominator, write the fraction as a sum like $\frac{A}{x - a} + \frac{B}{x - b}$ , solve for the constants, and integrate each piece, which usually gives logarithms. For example, $\frac{1}{x ( x + 1 )} = \frac{1}{x} - \frac{1}{x + 1}$ , so its integral is $ln ∣ x ∣ - ln ∣ x + 1∣ + C$ .

The formula

\frac{p x + q}{( x - a ) ( x - b )} = \frac{A}{x - a} + \frac{B}{x - b} ⟹ \int = A ln ∣ x - a ∣ + B ln ∣ x - b ∣ + C

Adding fractions, run in reverse. At school you were told to find a common denominator; now the job is to un-find one.

When to use it

Use partial fractions when the integrand is one polynomial divided by another and the bottom factors into pieces: $\frac{5 x + 1}{x ^{2} + x - 2}$ , $\frac{1}{x ^{2} - 4}$ , $\frac{x + 3}{x ( x + 1 ) ^{2}}$ . Before you start, check two things. First, is the top the derivative of the bottom (or a multiple of it)? Then u-substitution gives $ln ∣ bottom ∣$ in one line. Second, is the degree of the top at least the degree of the bottom? Then do polynomial long division first, and use partial fractions on the remainder.

How to do it, step by step

Factor the denominator completely. $x^{2} + x - 2 = (x - 1) (x + 2)$ .
Write the split with unknown constants. One $\frac{A}{x - a}$ for each distinct linear factor. A repeated factor $(x - a)^{2}$ gets two terms, $\frac{A}{x - a} + \frac{B}{( x - a ) ^{2}}$ .
Multiply through by the denominator. You get a polynomial identity, like $5 x + 1 = A (x + 2) + B (x - 1)$ , that holds for every $x$ .
Find the constants. Substitute each root of the denominator in turn: each one makes every term but one vanish. If you run out of roots, compare coefficients of a power of $x$ .
Integrate each piece. $\int \frac{A}{x - a} d x = A ln ∣ x - a ∣$ and $\int \frac{B}{( x - a ) ^{2}} d x = - \frac{B}{x - a}$ . Add $+ C$ at the end.

Worked examples

Example 1: The classic split

Easy

Integrate:

\int \frac{1}{x ( x + 1 )} d x

The bottom is already factored. Split it:

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

Put $x = 0$ : $1 = A$ . Put $x = - 1$ : $1 = - B$ , so $B = - 1$ .

\int \frac{1}{x ( x + 1 )} d x = \int (\frac{1}{x} - \frac{1}{x + 1}) d x

Using the log laws, you can also write the answer as $ln \frac{x}{x + 1} + C$ . The same split, $\frac{1}{n} - \frac{1}{n + 1}$ , is what makes the series $\sum \frac{1}{n ( n + 1 )}$ collapse to 1, so it is worth knowing by heart.

Answer

\int \frac{1}{x ( x + 1 )} d x = ln ∣ x ∣ - ln ∣ x + 1∣ + C

Example 2: Factor first

Medium

Integrate, for $x > 1$ :

\int \frac{5 x + 1}{x ^{2} + x - 2} d x

The top is not a multiple of the bottom's derivative ( $2 x + 1$ ), and its degree is lower, so partial fractions it is. Factor: $x^{2} + x - 2 = (x - 1) (x + 2)$ .

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

Put $x = 1$ : $6 = 3 A$ , so $A = 2$ . Put $x = - 2$ : $- 9 = - 3 B$ , so $B = 3$ .

\int \frac{5 x + 1}{x ^{2} + x - 2} d x = \int (\frac{2}{x - 1} + \frac{3}{x + 2}) d x

Check it by adding the fractions back: $\frac{2 ( x + 2 ) + 3 ( x - 1 )}{( x - 1 ) ( x + 2 )} = \frac{5 x + 1}{x ^{2} + x - 2}$ . It works.

Answer

\int \frac{5 x + 1}{x ^{2} + x - 2} d x = 2 ln ∣ x - 1∣ + 3 ln ∣ x + 2∣ + C

Example 3: A repeated factor

Harder

Integrate, for $x > 0$ :

\int \frac{x + 3}{x ( x + 1 ) ^{2}} d x

The factor $(x + 1)$ appears twice, so it needs two terms. Call the constants $A$ , $B$ and $D$ , to keep $C$ free for the end.

\frac{x + 3}{x ( x + 1 ) ^{2}} = \frac{A}{x} + \frac{B}{x + 1} + \frac{D}{( x + 1 ) ^{2}}

x + 3 = A (x + 1)^{2} + B x (x + 1) + D x

Put $x = 0$ : $3 = A$ . Put $x = - 1$ : $2 = - D$ , so $D = - 2$ . There are no roots left for $B$ , so compare the $x^{2}$ terms: the left has none, the right has $A + B$ . So $A + B = 0$ and $B = - 3$ .

\int (\frac{3}{x} - \frac{3}{x + 1} - \frac{2}{( x + 1 ) ^{2}}) d x

The first two are logs. The last one is a power, $- 2 (x + 1)^{- 2}$ , which integrates to $+ 2 (x + 1)^{- 1}$ . Not a log: that is the trap with repeated factors.

Answer

\int \frac{x + 3}{x ( x + 1 ) ^{2}} d x = 3 ln ∣ x ∣ - 3 ln ∣ x + 1∣ + \frac{2}{x + 1} + C

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Integrate, for $x > 0$ :

\int \frac{6}{x ^{2} + 3 x} d x

Problem 2

Medium

Integrate, for $x > 2$ :

\int \frac{x + 7}{x ^{2} - x - 2} d x

Problem 3

Harder

Integrate, for $x > 1$ :

\int \frac{x ^{2} + 1}{x ^{2} - 1} d x

Common mistakes

Using it when substitution would do. $\int \frac{2 x + 1}{x ^{2} + x - 2} d x$ splits into partial fractions, but the top is exactly the derivative of the bottom, so the answer is just $ln ∣ x^{2} + x - 2∣ + C$ . Check for that first; it takes two seconds.
Skipping the long division. If the top's degree is at least the bottom's, the split does not work. Divide first: $\frac{x ^{2}}{x ^{2} - 1} = 1 + \frac{1}{x ^{2} - 1}$ .
One term for a repeated factor. $(x + 1)^{2}$ needs both $\frac{B}{x + 1}$ and $\frac{D}{( x + 1 ) ^{2}}$ . With only one, the constants will not solve.
Turning every piece into a log. $\int \frac{1}{( x + 1 ) ^{2}} d x = - \frac{1}{x + 1}$ , not $ln ∣ x + 1 ∣^{2}$ .
Losing a coefficient inside the log. $\int \frac{1}{2 x + 1} d x = \frac{1}{2} ln ∣2 x + 1∣$ , because of the chain rule. A plain $\frac{1}{x - a}$ needs no adjustment, but $\frac{1}{k x - a}$ does.

Practice it now

No clock, three strikes

Partial fractions only appear in Diverge, from about difficulty 7 or 8: a linear top over a quadratic with two different linear factors, so the answer is always one or two logarithms. Repeated factors like Example 3 are not in the game yet. The $+ C$ is optional when you type it.

Play Diverge See it on the formula sheet

FAQ

When do you use partial fractions?: When you are integrating a rational function whose bottom factors, and substitution does not work directly. It is the standard method for things like $\int \frac{1}{x ^{2} - 1} d x$ .
What if the denominator does not factor?: A quadratic like $x^{2} + 4$ that will not factor gets a term $\frac{A x + B}{x ^{2} + 4}$ , which integrates to a mix of $ln (x^{2} + 4)$ and $arctan \frac{x}{2}$ . The arctan part is the territory of trig substitution.
Is there a shortcut for finding the constants?: Yes, the cover-up method. For a distinct factor $x - a$ , cover it up in the original fraction and substitute $x = a$ into what is left. For $\frac{5 x + 1}{( x - 1 ) ( x + 2 )}$ , covering $(x - 1)$ and putting $x = 1$ gives $\frac{6}{3} = 2$ , which is $A$ . It does not work for the lower terms of repeated factors.
Why are there absolute values in the answer?: Because $ln$ only takes positive inputs, and $\int \frac{1}{u} d u = ln ∣ u ∣ + C$ is valid on both sides of $u = 0$ . On an interval where $x - 1 > 0$ you can drop them.
Where can I find partial fractions practice problems?: Right on this page: "Your turn" has three partial fractions practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, Diverge, which has no clock, is free and needs no account.

Partial Fractions: Steps, Examples & Practice Problems

$An illustration of Partial fractions$

When to use it

Use partial fractions when the integrand is one polynomial divided by another and the bottom factors into pieces:

\frac{5 x + 1}{x ^{2} + x - 2}

\frac{1}{x ^{2} - 4}

\frac{x + 3}{x ( x + 1 ) ^{2}}

. Before you start, check two things. First, is the top the derivative of the bottom (or a multiple of it)? Then u-substitution gives

ln ∣ bottom ∣

in one line. Second, is the degree of the top at least the degree of the bottom? Then do polynomial long division first, and use partial fractions on the remainder.

How to do it, step by step

Factor the denominator completely.

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

Write the split with unknown constants. One

\frac{A}{x - a}

for each distinct linear factor. A repeated factor

(x - a)^{2}

gets two terms,

\frac{A}{x - a} + \frac{B}{( x - a ) ^{2}}

Multiply through by the denominator. You get a polynomial identity, like

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

, that holds for every

x

Find the constants. Substitute each root of the denominator in turn: each one makes every term but one vanish. If you run out of roots, compare coefficients of a power of

x

Integrate each piece.

\int \frac{A}{x - a} d x = A ln ∣ x - a ∣

and

\int \frac{B}{( x - a ) ^{2}} d x = - \frac{B}{x - a}

. Add

+ C

at the end.

Worked examples

Example 1: The classic split

Easy

Integrate:

\int \frac{1}{x ( x + 1 )} d x

The bottom is already factored. Split it:

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

Put $x = 0$ : $1 = A$ . Put $x = - 1$ : $1 = - B$ , so $B = - 1$ .

\int \frac{1}{x ( x + 1 )} d x = \int (\frac{1}{x} - \frac{1}{x + 1}) d x

Answer

\int \frac{1}{x ( x + 1 )} d x = ln ∣ x ∣ - ln ∣ x + 1∣ + C

Example 2: Factor first

Medium

Integrate, for $x > 1$ :

\int \frac{5 x + 1}{x ^{2} + x - 2} d x

The top is not a multiple of the bottom's derivative ( $2 x + 1$ ), and its degree is lower, so partial fractions it is. Factor: $x^{2} + x - 2 = (x - 1) (x + 2)$ .

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

Put $x = 1$ : $6 = 3 A$ , so $A = 2$ . Put $x = - 2$ : $- 9 = - 3 B$ , so $B = 3$ .

\int \frac{5 x + 1}{x ^{2} + x - 2} d x = \int (\frac{2}{x - 1} + \frac{3}{x + 2}) d x

Check it by adding the fractions back: $\frac{2 ( x + 2 ) + 3 ( x - 1 )}{( x - 1 ) ( x + 2 )} = \frac{5 x + 1}{x ^{2} + x - 2}$ . It works.

Answer

\int \frac{5 x + 1}{x ^{2} + x - 2} d x = 2 ln ∣ x - 1∣ + 3 ln ∣ x + 2∣ + C

Example 3: A repeated factor

Harder

Integrate, for $x > 0$ :

\int \frac{x + 3}{x ( x + 1 ) ^{2}} d x

The factor $(x + 1)$ appears twice, so it needs two terms. Call the constants $A$ , $B$ and $D$ , to keep $C$ free for the end.

\frac{x + 3}{x ( x + 1 ) ^{2}} = \frac{A}{x} + \frac{B}{x + 1} + \frac{D}{( x + 1 ) ^{2}}

x + 3 = A (x + 1)^{2} + B x (x + 1) + D x

\int (\frac{3}{x} - \frac{3}{x + 1} - \frac{2}{( x + 1 ) ^{2}}) d x

The first two are logs. The last one is a power, $- 2 (x + 1)^{- 2}$ , which integrates to $+ 2 (x + 1)^{- 1}$ . Not a log: that is the trap with repeated factors.

Answer

\int \frac{x + 3}{x ( x + 1 ) ^{2}} d x = 3 ln ∣ x ∣ - 3 ln ∣ x + 1∣ + \frac{2}{x + 1} + C

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Integrate, for $x > 0$ :

\int \frac{6}{x ^{2} + 3 x} d x

Problem 2

Medium

Integrate, for $x > 2$ :

\int \frac{x + 7}{x ^{2} - x - 2} d x

Problem 3

Harder

Integrate, for $x > 1$ :

\int \frac{x ^{2} + 1}{x ^{2} - 1} d x

Common mistakes

Using it when substitution would do.

\int \frac{2 x + 1}{x ^{2} + x - 2} d x

splits into partial fractions, but the top is exactly the derivative of the bottom, so the answer is just

ln ∣ x^{2} + x - 2∣ + C

. Check for that first; it takes two seconds.

Skipping the long division. If the top's degree is at least the bottom's, the split does not work. Divide first:

\frac{x ^{2}}{x ^{2} - 1} = 1 + \frac{1}{x ^{2} - 1}

One term for a repeated factor.

(x + 1)^{2}

needs both

\frac{B}{x + 1}

and

\frac{D}{( x + 1 ) ^{2}}

. With only one, the constants will not solve.

Turning every piece into a log.

\int \frac{1}{( x + 1 ) ^{2}} d x = - \frac{1}{x + 1}

, not

ln ∣ x + 1 ∣^{2}

Losing a coefficient inside the log.

\int \frac{1}{2 x + 1} d x = \frac{1}{2} ln ∣2 x + 1∣

, because of the chain rule. A plain

\frac{1}{x - a}

needs no adjustment, but

\frac{1}{k x - a}

does.

Practice it now

No clock, three strikes

Play Diverge See it on the formula sheet

FAQ

When do you use partial fractions?

When you are integrating a rational function whose bottom factors, and substitution does not work directly. It is the standard method for things like

\int \frac{1}{x ^{2} - 1} d x

What if the denominator does not factor?

A quadratic like

x^{2} + 4

that will not factor gets a term

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

, which integrates to a mix of

ln (x^{2} + 4)

and

arctan \frac{x}{2}

. The arctan part is the territory of trig substitution.

Is there a shortcut for finding the constants?

Yes, the cover-up method. For a distinct factor

x - a

, cover it up in the original fraction and substitute

x = a

into what is left. For

\frac{5 x + 1}{( x - 1 ) ( x + 2 )}

, covering

(x - 1)

and putting

x = 1

gives

\frac{6}{3} = 2

, which is

A

. It does not work for the lower terms of repeated factors.

Why are there absolute values in the answer?

Because

ln

only takes positive inputs, and

\int \frac{1}{u} d u = ln ∣ u ∣ + C

is valid on both sides of

u = 0

. On an interval where

x - 1 > 0

you can drop them.

Where can I find partial fractions practice problems?

Right on this page: "Your turn" has three partial fractions practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, Diverge, which has no clock, is free and needs no account.

Partial Fractions: Steps, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: The classic split

Example 2: Factor first

Example 3: A repeated factor

Your turn

Common mistakes

FAQ

Keep going

Partial Fractions: Steps, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: The classic split

Example 2: Factor first

Example 3: A repeated factor

Your turn

Common mistakes

FAQ

Keep going