Partial fraction

1. 2016
Submitted To: MS
Quratul Ain
Submitted By:
Muhammad Saad
University of Gujrat(Marghzar Campus)

2. University of Gujrat
Page 1
Contents
Topics page no.
Partial Fraction 2
Examples of partial Fraction
 Example1 2
 Example2 3
 Example3 3
Application of Partial Fraction 4
Reference 5

3. University of Gujrat
Page 2
Assignment of Calculus
Partial Fractions
When a proper rational expression is decomposed into a sum of two or more rational
expressions, it is known as Partial Fractions.
It is used in integrating rational fractions in calculus and finding the inverse Laplace transform.
In partial fractions the degree of numerator is less than the degree of the denominator. [1]
Examples of PartialFractions
Example 1
The rational function
𝑥−4
𝑥(𝑥+4)
can be decomposed into partial fractions in the following way:
First decompose the fraction into linear factors as
𝑥 − 4
𝑥(𝑥 + 4)
=
𝐴
𝑥
+
𝐵
(𝑥 + 4)
On simplification, x-4 = A(x+4) + B(x)
Now, by comparing the co-efficient of like terms on both sides, we get,
A+B = 1, 4 A = -4
On solving equations, we get, A = -1, B = 2.
By substituting the value of A and B we get,
𝑥 − 4
𝑥( 𝑥 + 4)
= −
1
𝑥
+
2
(𝑥 + 4)

4. University of Gujrat
Page 3
Example 2
The rational function
1
𝑥2+2𝑥+3
can be decomposed into partial fractions in the following way:
First decompose the fraction into linear factors as [2]
1
𝑥2 + 2𝑥 − 3
=
𝐴
𝑥 + 3
+
𝐵
𝑥 − 1
On simplification, 1 = A(x-1) + b(x+3)
Now, by comparing the co-efficient of like terms on both sides, we get,
A+B = 1, -4A = 1
On solving equations, we get, A = −
1
4
, 𝐵 =
1
4
By substituting the value of A and B we get,
1
𝑥2 + 2𝑥 − 3
= −
1
4( 𝑥 + 3)
+
1
4( 𝑥 − 1)
Example 3
The rational function
1
𝑥2−𝑥−6
can be decomposed into partial fractions in the following way:
First decompose the fraction into linear factors as
1
𝑥2−𝑥−6
=
𝐴
𝑥−3
+
𝐵
𝑥+2
On simplification, 1=A(x+2) + B(x-3)

5. University of Gujrat
Page 4
Now, by comparing the co-efficient of like terms on both sides, we get,
A+B = 1, 1 = 5A
On solving equations, we get, A =
1
5
, B = -
1
5
By substituting the value of A and B we get,
1
𝑥2 −𝑥−6
=
1
5
[(𝑥 − 3) − (𝑥 + 2)]
Application of Partial Fraction
Systems of linear equations are used in the real world by economists to find out when supply
equals demand, and if you don’t know the numbers when you have a business, it might fail.
Whoever said life was linear? Most real-life equations are actually non-linear, like throwing a
ball up high. Picture yourself throwing a tennis ball into the air, then one or two seconds later,
another ball you could use a system of non-linear equations, in this case quadratic equations, to
find out when two balls would be the same height.
An abstract concept we covered in this section involved partial fractions. This topic seems
unconnected with real life, but we can really use this in calculus. In calculus, we get area under
curves by a method called integration. When finding the area of a crazy algebraic fraction, it
helps to break the fraction into smaller chunks to make integration easier.
Why would we need to find the area under a curve? That’s a whole other topic, but you might a
painting a huge mural on the side of a building and need to calculate the painted area of a curve
to determine the amount of paint to buy. Depending on the complexity of the curve, you might
have to use partial fractions to help you integrate. There, now you should feel better about partial
fractions. [3]

6. University of Gujrat
Page 5
Reference
[1] www.icoachmath.com
[2] https://www.boundless.com
[3] www.shmoop.com