7.4 Integration by Partial Fractions
The method of partial fractions is used to integrate rational functions. That is, we want to compute
P( x )
Z
dx where P, Q are polynomials.
Q( x )
R( x )
First reduce1 the integrand to the form S( x ) + Q( x )
where °R < °Q.
Example Here we write the integrand as a polynomial plus a rational function x+7 2 whose denom-
inator has higher degreee than its numerator. Thankfully, this expression can be easily integrated
using logarithms.
x2 + 3 x ( x + 2) − 2x + 3 −2( x + 2) + 4 + 3 7
= = x+ = x−2+
x+2 x+2 x+2 x+2
x2 + 3 7 1
Z Z
=⇒ dx = x−2+ dx = x2 − 2x + 7 ln | x + 2| + c
x+2 x+2 2
What if °Q ≥ 2?
If the denominator Q( x ) is quadratic or has higher degree, we need another trick:
R( x )
Theorem. Suppose that °R < °Q. Then the rational function Q( x )
can be written as a sum of fractions of the
form
A Ax + B
( ax + b)m ( ax2+ bx + c)n
where A, B, a, b, c are constants and m, n are positive integers.
Expressions such as the above can all be integrated using either logarithms or trigonometric substi-
tutions.
Example With a little experimenting, you should be convinced that
3x2 + 2x + 3 3 2
3
= +
x +x x 1 + x2
It follows that
3x2 + 2x + 3
Z
dx = 3 ln | x | + 2 tan−1 x + c
x3 + x
The burning question is how to find the expressions in the Therorem. The approach depends on
the form of the denominator Q( x ).
1 By Long Division or some other Torture. . .
1
, Case 1: Distinct Linear Factors
Suppose that our denominator can be factorized completely into distinct linear factors. That is
Q( x ) = ( x − a1 )( x − a2 ) · · · ( x − an )
where the values a1 , . . . , an are all different.2
Theorem. For such a Q, there exist constants A1 , . . . , An such that
n
R( x ) Ai A1 An
=∑ = +···+ (∗)
Q( x ) i =1
x − a i x − a 1 x − an
whence the integral can be easily computed term-by-term:
n n
R( x ) Ai
Z Z
dx = ∑ dx = ∑ Ai ln | x − ai | + c
Q( x ) i =1
x − ai i =1
We find the constants Ai by putting the right hand side of (∗) over the common denominator Q( x )
R( x ) R( x ) A1 An
= = +···+
Q( x ) ( x − a1 ) · · · ( x − a n ) x − a1 x − an
and comparing numerators.
Examples
1. According to the Theorem, there exist constants A, B such that
x+8 x+8 A B
= = +
x2 +x−2 ( x − 1)( x + 2) x−1 x+2
Summing the right hand side, we obtain
x+8 A ( x + 2) + B ( x − 1)
=
( x − 1)( x + 2) ( x − 1)( x + 2)
Since the denominators are equal, it follows that the numerators are equal:
x + 8 = A ( x + 2) + B ( x − 1)
This is a relationship between A, B which holds for all3 x: every value of x gives a valid rela-
tionship between A and B. Evaluating at x = 1 and x = −2 gives two very simple expressions:
x=1: 9 = 3A =⇒ A = 3
x = −2 : 6 = −3B =⇒ B = −2
Putting it all together, we have
x+8 3 2
Z Z
2
dx = − dx = 3 ln | x − 1| − 2 ln | x + 2| + c
x +x−2 x−1 x+2
| x − 1|3
= ln +c
| x + 2|2
2 We assume for clarity that the leading term of Q( x ) is x n (coefficient 1). If not, absorb it into the numerator!
3 You might worry that it doesn’t when x = 1 or x = −2 because of the denominator. The fact fact that polynomials are
continuous combined with x + 8 = A( x + 2) + B( x − 1) everywhere else guarantees that we have equality everywhere.
2
The method of partial fractions is used to integrate rational functions. That is, we want to compute
P( x )
Z
dx where P, Q are polynomials.
Q( x )
R( x )
First reduce1 the integrand to the form S( x ) + Q( x )
where °R < °Q.
Example Here we write the integrand as a polynomial plus a rational function x+7 2 whose denom-
inator has higher degreee than its numerator. Thankfully, this expression can be easily integrated
using logarithms.
x2 + 3 x ( x + 2) − 2x + 3 −2( x + 2) + 4 + 3 7
= = x+ = x−2+
x+2 x+2 x+2 x+2
x2 + 3 7 1
Z Z
=⇒ dx = x−2+ dx = x2 − 2x + 7 ln | x + 2| + c
x+2 x+2 2
What if °Q ≥ 2?
If the denominator Q( x ) is quadratic or has higher degree, we need another trick:
R( x )
Theorem. Suppose that °R < °Q. Then the rational function Q( x )
can be written as a sum of fractions of the
form
A Ax + B
( ax + b)m ( ax2+ bx + c)n
where A, B, a, b, c are constants and m, n are positive integers.
Expressions such as the above can all be integrated using either logarithms or trigonometric substi-
tutions.
Example With a little experimenting, you should be convinced that
3x2 + 2x + 3 3 2
3
= +
x +x x 1 + x2
It follows that
3x2 + 2x + 3
Z
dx = 3 ln | x | + 2 tan−1 x + c
x3 + x
The burning question is how to find the expressions in the Therorem. The approach depends on
the form of the denominator Q( x ).
1 By Long Division or some other Torture. . .
1
, Case 1: Distinct Linear Factors
Suppose that our denominator can be factorized completely into distinct linear factors. That is
Q( x ) = ( x − a1 )( x − a2 ) · · · ( x − an )
where the values a1 , . . . , an are all different.2
Theorem. For such a Q, there exist constants A1 , . . . , An such that
n
R( x ) Ai A1 An
=∑ = +···+ (∗)
Q( x ) i =1
x − a i x − a 1 x − an
whence the integral can be easily computed term-by-term:
n n
R( x ) Ai
Z Z
dx = ∑ dx = ∑ Ai ln | x − ai | + c
Q( x ) i =1
x − ai i =1
We find the constants Ai by putting the right hand side of (∗) over the common denominator Q( x )
R( x ) R( x ) A1 An
= = +···+
Q( x ) ( x − a1 ) · · · ( x − a n ) x − a1 x − an
and comparing numerators.
Examples
1. According to the Theorem, there exist constants A, B such that
x+8 x+8 A B
= = +
x2 +x−2 ( x − 1)( x + 2) x−1 x+2
Summing the right hand side, we obtain
x+8 A ( x + 2) + B ( x − 1)
=
( x − 1)( x + 2) ( x − 1)( x + 2)
Since the denominators are equal, it follows that the numerators are equal:
x + 8 = A ( x + 2) + B ( x − 1)
This is a relationship between A, B which holds for all3 x: every value of x gives a valid rela-
tionship between A and B. Evaluating at x = 1 and x = −2 gives two very simple expressions:
x=1: 9 = 3A =⇒ A = 3
x = −2 : 6 = −3B =⇒ B = −2
Putting it all together, we have
x+8 3 2
Z Z
2
dx = − dx = 3 ln | x − 1| − 2 ln | x + 2| + c
x +x−2 x−1 x+2
| x − 1|3
= ln +c
| x + 2|2
2 We assume for clarity that the leading term of Q( x ) is x n (coefficient 1). If not, absorb it into the numerator!
3 You might worry that it doesn’t when x = 1 or x = −2 because of the denominator. The fact fact that polynomials are
continuous combined with x + 8 = A( x + 2) + B( x − 1) everywhere else guarantees that we have equality everywhere.
2
