Integrating Rational Functions by Partial Fractions
Using Linear Factors
One of the strategies used extensively as a method of integration involves using partial fractions. For instance, try to integrate the following:
dx
∫ x2 + x − 2
This equation cannot be integrated as is. For example, a direct "u" substitution cannot be made because the corresponding dx's would not match. (Example
shown below).
Let u = x 2 + x − 2
Then du = x − 2
The original problem is missing the (2x+1) portion.
Instead of substitution, partial fractions can be used. This method breaks down the original equation into several equations that can be integrated easily.
This handout will demonstrate the basic steps.
Step 1: Check to make sure the original equation is a proper rational function.
When working with partial fractions, the original equation must be written as a proper rational function. This means that the degree of the
numerator is less than the degree of the denominator. In other words, if the numerator contains an "x3" and the denominator contains an "x2", then the
fraction must be reduced. This is done by either factoring and eliminating terms or by performing long division so that the smaller power will rest in the
numerator. In the example given at the beginning, the numerator already contains the smaller power of x: x0.
Step 2: Factor the polynomial in the denominator into linear factors.
In the example, the denominator contains a quadratic polynomial; therefore, before beginning with the partial fraction decomposition, we must
first factor the quadratic into its linear factors. (refer to our handout for factoring polynomials)
x 2 + x − 2 = ( x − 1)( x + 2)
Step 3: Set up the partial fraction decomposition.
The general form for linear factors is as follows:
Am
A1
A2
. Therefore, the partial fraction decomposition
+
+ ... +
2
ax + b (ax + b)
(ax + b) m
for the original equation would look like this:
1
1
A
B
=
=
+
x + x − 2 ( x − 1)( x + 2) x − 1 x + 2
2
Step 4: Identify A and B.
•
Begin by multiplying through to get rid of the fractions (multiply everything by the common denominator)
( x − 1)( x + 2)[
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1
A
B
=
+
] => 1 = A( x + 2) + B( x − 1)
( x − 1)( x + 2) x − 1 x + 2
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•
Multiply the A's and B's through in order to group common terms
1 = Ax + 2 A + Bx − B => 1 = ( A + B) x + (2 A − B)
•
Set up a system of equations to find the values of A and B. This is done by writing equations that group all the x terms in one equation and all the
constants in another.
A+ B = 0
2A − B = 1
Note: Since there is no corresponding coefficient for the first term, x, on the left-hand side, we can say that
the coefficients of x on the right-hand side are equal to zero.
•
Solve the system of equations using the elimination or addition method. (refer to our handout on solving systems of equations)
A+B=0
+2A - B = 1
3A = 1
A = 1/3
Adding the two equations eliminates the B term
and leaves the A term which can be solved for.
Plug the value for A back into one of the original equations to find B:
A + B = 0 => 1/3 + B = 0 => B= -1/3
•
Use the values for A and B to re-write the original integral
∫x
2
− 1/ 3
dx
1/ 3
=∫
⋅ dx + ∫
⋅ dx
x −1
x+2
+x−2
=>
∫x
2
dx
dx
dx
=∫
−∫
3( x − 1)
3( x + 2)
+x−2
Step 5: Integrate
dx
dx
dx
dx
∫ 3( x − 1) − ∫ 3( x + 2) = > ∫ x − 1 − ∫ x + 2 = >
1
3
1
3
1
3
ln x − 1 − 13 ln x + 2 + c
So,
∫x
2
dx
=
+ x−2
1
3
ln x − 1 − 1 3 ln x + 2 + c or
1
3
ln
x −1
+c
x+2
(Integrating Rational Functions by Partial Fractions Using Linear Factors continued (2))
For problems that involve quadratic factors (i.e. in the form ax 2 + bx + c ) a different strategy must be used.
The general form for Quadratic Factors is:
Am x + Bm
A1 x + B1
A2 x + B2
+
+ ... +
2
2
2
ax + bx + c (ax + bx + c)
(ax 2 + bx + c) m
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Notice how the term on the numerator changes for the decomposition of quadratic factors when compared to linear factors.
Example 2
x 2 + 4 x − 23
x 3 + 3 x 2 + 4 x + 12
Decompose the following:
x 2 + 4 x − 23
x 2 + 4 ( x + 3)
(
First factor the denominator. (refer to our hand out on factoring by grouping). Now the equation should look like:
)
Now, in the denominator there is a quadratic factor and a linear factor so a combination of both methods should be used. The setup for the decomposition
should look like:
Ax + B
C
x 2 + 4 x − 23
=
+
(x
2
)
+ 4 ( x + 3)
x2 + 4
x+3
Now, following the strategy of the previous example, multiply both sides by the common denominator as shown:
(x
2
)
+ 4 ( x + 3) [
x 2 + 4 x − 23
C
Ax + B
= 2
] =>
+
2
x + 4 ( x + 3) x + 4 x + 3
(
)
(
x 2 + 4 x − 23 = ( Ax + B )( x + 3) + C x 2 + 4
)
Multiply everything out and group together like terms.
x 2 + 4 x − 23 = Ax 2 + 3 Ax + Bx + 3B + Cx 2 + 4C =>
x 2 + 4 x − 23 = Ax 2 + Cx 2 + 3 Ax + Bx + 3B + 4C
Now separate the common terms into separate equations as shown. (i.e. all the x 2 terms in one equation, all the x terms in another, and all the constants
in another.)
x 2 = Ax 2 + Cx 2
4 x = 3 Ax + Bx
− 23 = 3B + 4C
For the first equation, divide both sides by x 2 , and for the second equation divide both sides by x to get:
1) 1 = A + C
2) 4 = 3 A + B
3) − 23 = 3B + 4C
Now there is a system of equations that must be solved by either elimination or substitution method. (refer to our hand out on solving systems of
equations.)
Take the first equation and subtract A from both sides to get
4) 1 − A = C
Take the second and subtract 3A from both sides to get
5) 4 − 3 A = B
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Now plug these solutions into the third equation and solve for A as follows.
− 23 = 3(4 − 3 A) + 4(1 − A)
Multiply Through and add to get:
− 23 = −13 A + 16
Subtract 16 from both sides
− 39 = −13 A
Divide by -13
3= A
Now, plug the value found for A into the equations marked 4) and 5) to get
1− 3 = C
−2=C
And
4 − 3(3) = B
−5 = B
Thus the system yields A=3, B=-5 and C=-2.
Now going back to the beginning, the decomposition of the fraction should be as follows:
x 2 + 4 x − 23
C
Ax + B
= 2
=>
+
2
x + 4 ( x + 3) x + 4 x + 3
(
)
x 2 + 4 x − 23
3x − 5 − 2
+
= 2
2
x + 4 ( x + 3) x + 4 x + 3
(
)
Remember, the main advantage of partial fractions is to transform a difficult integral problem into a series of equations with easier integrals.
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