Partial Fraction Decomposition

How to use the partial fraction decomposition to simplify a rational expression: formula, 3 examples, and their solutions.

Example 1

Example

Solution

The partial fraction decomposition is a way
to split a rational expression
to its partial fractions.
(= fractions with reduced denominators)

See the given expression.
The denominator is x(x - 1).

The reduced factors are
x and (x - 1).

So set
(given) = A/x + B/(x - 1).

The goal is to find A and B.

Change the denominators
to x(x - 1).

Add and Subtract Rational Expressions

See A/x.

The denominator is x.

The factor (x - 1) is missing.

So multiply (x - 1)
to both of the numerator and the denominator.

A/x
= [A/x]⋅[(x - 1)/(x - 1)]

Next, see +B/(x - 1).

The denominator is (x - 1).

The factor x is missing.

So multiply x
to both of the numerator and the denominator.

+B/(x - 1)
= [B/(x - 1)]⋅[x/x]

So
A/x + B/(x - 1)
= [A/x]⋅[(x - 1)/(x - 1)] + [B/(x - 1)]⋅[x/x].

A(x - 1) = Ax - A

Multiply a Monomial and a Polynomial

+B⋅x = +Bx

Ax + Bx = (A + B)x

Common Monomial Factor

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

The denominators are the same.
So the numerators are also the same.

The x terms are the same.
So the coefficients are the same.

So A + B = 3.

The constant terms are the same.

So -A = -2.

A + B = 3
-A = -2

Find A and B
by solving this system.

-A = -2

So A = 2.

Put A = 2
into A + B = 3.

Then 2 + B = 3.

Substitution Method

Move 2 to the right side.

Then B = 1.

(given) = A/x + B/(x - 1)

A = 2
B = 1

So (given) = 2/x + 1/(x - 1).

So
2/x + 1/(x - 1)
is the answer.

Example 2

Example

Solution

See the given expression.
The denominator is x2(x + 1).

The reduced factors are
x, x2, and (x + 1).

So set
(given) = A/x + B/x2 + C/(x + 1).

The goal is to find A, B, and C.

Change the denominators
to x2(x + 1).

See A/x.

The denominator is x.

x(x + 1) is missing.

So multiply x(x + 1)
to both of the numerator and the denominator.

A/x
= [A/x]⋅[[x(x + 1)]/[x(x + 1)]]

See +B/x2.

The denominator is x2.

(x + 1) is missing.

So multiply (x + 1)
to both of the numerator and the denominator.

+B/x2
= +[B/x2]⋅[(x + 1)/(x + 1)]

See +C/(x + 1).

The denominator is (x + 1).

x2 is missing.

So multiply x2
to both of the numerator and the denominator.

+C/(x + 1)
= +[C/(x + 1)]⋅[x2/x2]

So
A/x + B/x2 + C/(x + 1)
= [A/x]⋅[[x(x + 1)]/[x(x + 1)]] + [B/x2]⋅[(x + 1)/(x + 1)] + [C/(x + 1)]⋅[x2/x2].

A⋅x(x + 1)
= Ax(x + 1)
= Ax2 + Ax

+B(x + 1) = +Bx + B

+C⋅x2 = +Cx2

Ax2 + Cx2 = (A + C)x2
Ax + Bx = (A + B)x

(5x2 - 1)/[x2(x + 1)]
= [(A + C)x2 + (A + B)x + B]/[x(x + 1)]

The denominators are the same.
So the numerators are also the same.

The x2 terms are the same.
So the coefficients are the same.

So A + C = 5.

There's no x term in the given expression.

So (A + B)x = 0x.

So A + B = 0.

The constant terms are the same.

So B = -1.

A + C = 5
A + B = 0
B = -1

Find A, B, and C
by solving this system.

System of Equations (3 Variables)

B = -1

One variable is already known.

Put B = -1
into A + B = 0.

Then A + (-1) = 0.

+(-1) = -1

Move -1 to the right side.

Then A = 1.

Put A = 1
into A + C = 5.

Then 1 + C = 5.

Move 1 to the right side.

Then C = 4.

(given) = A/x + B/x2 + C/(x + 1)

A = 1
B = -1
C = 4

So (given) = 1/x + (-1)/x2 + 4/(x + 1).

+(-1)/x2 = -1/x2

So
1/x - 1/x2 + 4/(x + 1)
is the answer.

Formula

Formula

1/AB = [1/(B - A)]⋅[1/A - 1/B]

Use this formula
when (B - A) doesn't have a variable.
(= constant)

Example 3

Example

Solution

The denominator is x(x + 1).

(x + 1) - x = 1
It doesn't have a variable.

So use the partial fraction decomposition formula.

1/[x(x + 1)] = [1/[(x + 1) - x]]⋅[1/x - 1/(x + 1)]

(x + 1) - x = 1

[1/1]⋅[1/x - 1/(x + 1)]
= 1/x - 1/(x + 1)

So
1/x - 1/(x + 1)
is the answer.