Partial Fractions for Integration: Factors and Solved Examples

A partial fractions question can look manageable until the denominator contains a square or a quadratic that refuses to factor. Suddenly you are wondering whether the numerator should be A, Ax + B, or something else entirely.

The reassuring part is that you do not have to guess. The denominator tells you the form to write. Once that form is correct, the rest is algebra followed by familiar integration.

This guide takes you from choosing the fractions to checking the final answer. We will work through distinct linear factors, repeated factors, irreducible quadratics, and an improper fraction that needs division first.

What are partial fractions?

Partial fractions rewrite one rational expression as a sum of simpler rational expressions. A rational expression is a ratio of polynomials.

For instance, combine these two fractions yourself:

2/(x - 1) + 3/(x + 2)
= [2(x + 2) + 3(x - 1)] / [(x - 1)(x + 2)]
= (5x + 1) / [(x - 1)(x + 2)]

Decomposition travels in the opposite direction. We start with the last expression and recover the two simpler ones.

Why bother? Because integrating 2/(x - 1) and 3/(x + 2) is straightforward. The combined fraction hides that simplicity.

Throughout this guide, ln means the natural logarithm, arctan means inverse tangent, and C is the final constant of integration. Formulae apply on intervals where the original integrand is defined.

Before splitting: simplify and compare degrees

First cancel any common polynomial factors, while retaining the original restrictions on x. Then compare the degrees of the numerator and denominator of the remaining fraction.

A proper fraction has a numerator degree smaller than its denominator degree. If the numerator degree is equal to or greater than the denominator degree, perform polynomial division first.

This degree check and the basic integration approach are covered in NCERT’s Integrals chapter, Section 7.5.

Also pause to check whether splitting is needed at all. For (2x + 3)/(x^2 + 3x + 7), the numerator is the derivative of the denominator, so direct substitution gives ln(x^2 + 3x + 7) + C. The quadratic is always positive.

If choosing the method is your main difficulty, start with our guide to substitution, parts and partial fractions.

How the denominator determines the setup

Here is a compact reference. Every letter in the right column represents an unknown constant.

Denominator factor	Terms to include
x - a	A/(x - a)
(x - a)^2	A/(x - a) + B/(x - a)^2
(x - a)^3	A/(x - a) + B/(x - a)^2 + D/(x - a)^3
Irreducible q(x) = ax^2 + bx + c	(Ax + B)/q(x)
Irreducible q(x)^2	(Ax + B)/q(x) + (Dx + E)/q(x)^2

Include every power up to the repeated factor’s highest power. Over a quadratic that cannot split into real linear factors, allow a numerator of degree at most one. Its x coefficient might eventually be zero, but do not assume that beforehand.

The complete factor-based setup is also set out in OpenStax’s partial fractions calculus section.

What does irreducible mean here?

We are factoring over the real numbers. A quadratic ax^2 + bx + c, with a nonzero, is irreducible over the reals when its discriminant b^2 - 4ac is negative.

• x^2 - 9 splits into (x - 3)(x + 3).
• x^2 + 6x + 9 is (x + 3)^2, a repeated linear factor.
• x^2 + 9 has a negative discriminant, so it stays quadratic.

Notice that x^2 itself is a repeated linear factor, not an irreducible quadratic. Its terms are A/x + B/x^2.

Example 1: distinct linear factors

Find the integral of:

(7x + 1) / [(x - 1)(x + 3)]

Write:

(7x + 1) / [(x - 1)(x + 3)]
= A/(x - 1) + B/(x + 3)

Multiply through by the common denominator:

7x + 1 = A(x + 3) + B(x - 1)

Choose values that remove one unknown at a time:

• At x = 1: 8 = 4A, so A = 2.
• At x = -3: -20 = -4B, so B = 5.

Therefore:

Integral = integral of [2/(x - 1) + 5/(x + 3)] dx
         = 2ln|x - 1| + 5ln|x + 3| + C

Check the algebra before moving on: 2(x + 3) + 5(x - 1) = 7x + 1, exactly the original numerator.

The excluded values are 1 and -3. An antiderivative is used on an interval that does not cross either value.

Why can we substitute a value that made the original denominator zero?

We substitute into the polynomial identity after clearing denominators, not into the original fraction. Once the numerator identity has been obtained, it holds as a polynomial identity, including at those helpful values. The original rational expression remains undefined there.

This distinction is small but useful: you are solving an algebraic identity, not assigning a value to division by zero.

Example 2: a repeated linear factor

Find the integral of:

(3x^2 + 3x - 3) / [(x - 1)^2(x + 2)]

The square means that both powers of x - 1 must appear:

= A/(x - 1) + B/(x - 1)^2 + D/(x + 2)

Clear the denominator:

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

At x = 1, the identity gives 3 = 3B, hence B = 1.

At x = -2, it gives 3 = 9D, hence D = 1/3.

Compare the coefficients of x^2: 3 = A + D, so A = 8/3.

Now integrate:

Integral = (8/3)ln|x - 1| - 1/(x - 1)
           + (1/3)ln|x + 2| + C

The middle term is a power-rule integral. Since 1/(x - 1)^2 = (x - 1)^(-2), its integral is -(x - 1)^(-1), not a logarithm.

A quick differentiation check is especially helpful here:

Derivative of -1/(x - 1) = +1/(x - 1)^2

Example 3: a linear factor and an irreducible quadratic

Find the integral of:

(3x^2 + x + 4) / [(x + 1)(x^2 + 4)]

Since x^2 + 4 cannot split over the reals, write:

= A/(x + 1) + (Bx + D)/(x^2 + 4)

After clearing denominators:

3x^2 + x + 4
= A(x^2 + 4) + (Bx + D)(x + 1)
= (A + B)x^2 + (B + D)x + (4A + D)

Match coefficients:

A + B = 3
B + D = 1
4A + D = 4

From the first two equations, B = 3 - A and D = A - 2. Put this into the third: 5A - 2 = 4, so:

A = 6/5, B = 9/5, D = -4/5

Split the quadratic numerator for integration:

Integral = (6/5)ln|x + 1|
           + (9/5) integral of x/(x^2 + 4) dx
           - (4/5) integral of 1/(x^2 + 4) dx

The first quadratic integral uses the derivative 2x. The second uses the inverse-tangent formula with a = 2:

Integral = (6/5)ln|x + 1|
           + (9/10)ln(x^2 + 4)
           - (2/5)arctan(x/2) + C

No absolute-value signs are needed around x^2 + 4 because it is positive for every real x.

This example shows why a constant numerator alone would have been too restrictive: the correct numerator contains both an x term and a constant.

Example 4: polynomial division comes first

Find the integral of:

(x^2 + 4)/(x^2 - 1)

Both degrees are 2. Divide first:

(x^2 + 4)/(x^2 - 1) = 1 + 5/(x^2 - 1)

Now factor and split the proper fraction:

5/[(x - 1)(x + 1)] = A/(x - 1) + B/(x + 1)
5 = A(x + 1) + B(x - 1)

At x = 1, A = 5/2. At x = -1, B = -5/2.

Therefore:

Integral = x + (5/2)ln|x - 1| - (5/2)ln|x + 1| + C

Do not lose the x from integrating the quotient 1. It is easy to become so absorbed in the fractions that the first part of the division disappears from the final answer.

Example 5: when the quadratic includes an x term

After decomposition, you may encounter:

Integral of (3x + 5)/(x^2 + 2x + 5) dx

The denominator’s derivative is 2x + 2. Rewrite the numerator to use it:

3x + 5 = (3/2)(2x + 2) + 2

Also complete the square:

x^2 + 2x + 5 = (x + 1)^2 + 4

Now the answer is:

(3/2)ln(x^2 + 2x + 5) + arctan((x + 1)/2) + C

To check the inverse-tangent coefficient, differentiate that term. Its derivative is 2/[(x + 1)^2 + 4], exactly the leftover part of the numerator.

The practical habit is to handle the derivative-shaped part first, then complete the square for what remains.

Extension: repeated irreducible quadratic factors

If the denominator contains (x^2 + 1)^2, the general contribution is:

(Ax + B)/(x^2 + 1) + (Dx + E)/(x^2 + 1)^2

You need both powers, each with its own possible linear numerator. This is the repeated-quadratic pattern described in OpenStax’s algebra treatment of partial fractions.

Here is a complete example with deliberately friendly coefficients:

Integral of (2x^3 + x^2 + 4x + 4)/(x^2 + 1)^2 dx

Clear the denominator in the proposed decomposition:

2x^3 + x^2 + 4x + 4
= (Ax + B)(x^2 + 1) + Dx + E

Comparing coefficients gives A = 2, B = 1, D = 2, E = 3. Hence the integrand is:

(2x + 1)/(x^2 + 1) + (2x + 3)/(x^2 + 1)^2

Most of these terms integrate immediately. The extra identity you need is:

Integral of 1/(x^2 + 1)^2 dx
= x/[2(x^2 + 1)] + (1/2)arctan x + C

You can verify it by differentiating the right side. Alternatively, substitute x = tan t; the integral becomes that of cos^2 t.

Combining the four contributions gives:

Integral = ln(x^2 + 1) + (5/2)arctan x
           - 1/(x^2 + 1) + 3x/[2(x^2 + 1)] + C

Treat this as an extension after the earlier examples feel comfortable. The setup follows the same pattern, but completing the integration can require an additional identity or substitution.

Three integration rules worth keeping beside your working

For nonzero a:

Integral of 1/(ax + b) dx = (1/a)ln|ax + b| + C

For an integer n > 1 and nonzero a:

Integral of (ax + b)^(-n) dx
= (ax + b)^(1 - n) / [a(1 - n)] + C

For positive a:

Integral of 1/(x^2 + a^2) dx = (1/a)arctan(x/a) + C

The coefficient outside a linear factor matters. For example, the integral of 4/(2x - 3) is 2ln|2x - 3| + C, not 4ln|2x - 3| + C.

Common mistakes and how to catch them

Mistake	A useful check
Writing only the highest repeated power	List powers from 1 upward before solving
Putting only a constant over an irreducible quadratic	Start with a possible linear numerator
Splitting an improper fraction immediately	Compare polynomial degrees first
Turning every term into a logarithm	Check whether the denominator has power 1
Missing the chain-rule coefficient	Differentiate the proposed answer
Dropping a quotient after division	Carry the polynomial part into the integral

A zero coefficient is not evidence that your setup was wrong. It simply means that the general form allowed a term that this particular numerator did not need.

For a definite integral, also inspect the interval before substituting limits. If a denominator vanishes inside it, ordinary endpoint subtraction is not justified. For example, the integral of 1/x from -1 to 1 is not an ordinary finite integral equal to zero; the two improper pieces diverge.

Try these before looking at the answers

For Questions 1 and 2, write only the decomposition form. For the remaining questions, integrate.

1. 1/[(x - 2)^3(x + 1)]
2. (x + 3)/[(x - 1)(x^2 + 2x + 5)^2]
3. (4x + 7)/[(x + 1)(x + 2)]
4. (x + 4)/(x + 1)^2
5. (2x^2 + 3)/(x^2 + 1)
6. (3x + 2)/(x^2 + 4)

Answers with the key step

1. Include every repeated power:

A/(x - 2) + B/(x - 2)^2 + D/(x - 2)^3 + E/(x + 1)

2. Use separate linear numerators over both quadratic powers:

A/(x - 1) + (Bx + D)/(x^2 + 2x + 5)
+ (Ex + F)/(x^2 + 2x + 5)^2

3. The decomposition is 3/(x + 1) + 1/(x + 2):

3ln|x + 1| + ln|x + 2| + C

4. Rewrite the numerator as (x + 1) + 3:

ln|x + 1| - 3/(x + 1) + C

5. Divide to get 2 + 1/(x^2 + 1):

2x + arctan x + C

6. Separate the derivative term from the constant term:

(3/2)ln(x^2 + 4) + arctan(x/2) + C

If an answer differs from yours, first compare the decompositions. A wrong coefficient is an algebra issue; a correct decomposition followed by a wrong antiderivative is an integration issue. Knowing which stage failed makes your next practice attempt much more useful.

Frequently asked questions

1. When should I use partial fractions for integration?

Use it when a rational function becomes easier to integrate after being split according to its denominator factors. Simplify and check for a direct substitution first. If the fraction is improper, divide before decomposing the remainder.

2. How do I handle distinct linear factors?

Assign one constant numerator to each factor, clear denominators, and solve for the constants. Roots of the factors are often convenient substitutions in the resulting polynomial identity.

3. Why must I include lower powers of a repeated factor?

The full expansion needs enough independent terms to represent a general numerator. For a cube such as (x - 2)^3, include denominators with powers 1, 2 and 3, even if a coefficient later becomes zero.

4. Why does an irreducible quadratic need Ax + B?

Its numerator must allow degree up to one. Restricting it to a constant may exclude the correct decomposition. Solve for both coefficients instead of guessing which one will vanish.

5. Is x squared an irreducible quadratic factor?

No. It is x multiplied by x, so use the repeated-linear form A/x + B/x^2.

6. Can I use the cover-up method for every coefficient?

The basic cover-up shortcut works directly for distinct linear factors. Substitution at a repeated root can isolate the highest-power term’s coefficient, but generally does not determine all the others. Use coefficient comparison or additional substitutions to finish.

7. Do all partial fraction integrals give logarithms?

No. Higher powers of linear factors produce power-rule terms, while irreducible quadratic terms may produce logarithms, inverse tangents, or further integrals requiring another technique.

8. Why do logarithms have absolute-value signs?

For a nonzero real expression u, differentiating ln|u| gives u'/u wherever it is defined. This handles intervals where u is negative too. If an expression is always positive, such as x^2 + 4, the bars can be omitted.

9. What is the quickest way to check my answer?

Recombine the fractions to check the algebra, then differentiate the antiderivative to check the calculus. These two checks catch different kinds of mistakes.

10. Do I need a separate constant after every term?

No. Combine additive constants into one C at the end of an indefinite integral. The chosen constant applies on the interval under consideration; intervals separated by an undefined point can have different constants.