Integration with Fractions
In This Chapter
- Completing the square
- Integration by partial fractions
- Working with nonrepeated linear factors
- Repeated linear factors
- Understanding irreducible quadratic factors
This is the last of the chapters on integration techniques. (Although Chapter 9 does deal with integration and issues with infinity.) In this chapter, we’ll study how to work with integrands that take particular fractional forms and transform them into manageable integrands. This would be a good time for you to look back at Chapter 1 and the process of dismantling fractional expressions.
We’ll finish the chapter with a number of integration problems that incorporate all we’ve studied in this book up to this point.
Completing the Square
Before we do some calculus, let’s take a minute to review a little bit of algebra. The expression (a + b)2 is clearly a square. The exponential 2 gives it away. When you expand (a + b)2 you get a2 + 2ab + b2. This is called a square trinomial. Get it? It’s a square and it has three terms. Observe that the first and third terms are squared, and the middle term is twice the product of the individual terms of the original binomial. The expression (2x + 3)2 = (2x)2 + 2(2x)(3) + (3)2 = 4x2 + 12x + 9. The expression (a – b)2 = a2 – 2ab + b2 is also a square trinomial.
The trinomial x2 + 8x + 9 is not a square trinomial. Although the first and third terms are squares, the middle term is not twice the product of x and 3. There is a technique called completing the square that is useful in algebra for creating equations in standard forms and in calculus for transforming an integrand into a recognizable integration form. See the upcoming examples to learn how to complete the square.
Critical Point
To complete the square on a quadratic ax2 + bx + c: Factor the leading coefficient from the first two terms of the quadratic to create the expression 
. Halve the linear term, 
, square it, add it within the parentheses, and subtract a times this value from c: 
. The expression is now 
.
Example 1: Complete the square with 4x2 + 12x + 10.
Solution:
1. Factor out the 4, 4(x2 + 3x) + 10.
2. Take half of 3, square it, and add the value inside the parentheses, while subtracting 4 times this amount from the 9:
3. Simplify this expression by writing the trinomial as a squared binomial and performing the arithmetic at the end of the expression:
Example 2: Evaluate 
.
Solution:
1. The denominator is the same as the trinomial from Example 1.
2. Completing the square on the denominator gives 
.
3. The denominator is the sum of two squares and can fit the form for the integral problem 
.
4. The question is what do we do about that 4?
5. Rewrite 
.
6. Let u = 2x + 3, which makes du = 2dx or 
.
7. Transform the integral to 
. Transforming back to the original variable, 
.
Example 3: Evaluate 
.
Solution:
1. Complete the square in the denominator: 3x2 + 6x + 4 = 3(x2 + 2x) + 4 = 3(x2 + 2x + 1) + 4 – 3.
2. Simplify this expression to be 3(x + 1)2 + 1. The integrand is now 
.
3. In order to perform the u-substitution, we need to move the 3 inside the parentheses. This is not going to be as “clean” as moving the 4 inside the parentheses in Example 2:
4. Let 
, so 
and 
.
5. Transform the integral to 
.
6. Working back to the original variable, 
.
Example 4: Evaluate 
.
Solution:
1. Complete the square in the denominator: x2 + 8x + 20 = (x2 + 8x + 16) + 4 = (x + 4)2 + 4.
2. Rewrite the integrand: 
. The form of the integrand for tan–1(u) is of the form 
, not 
. We need to factor out the 4 as follows:
3. Let 
so that 
or that dx = 2du:
4. Transform the result back to the original variable, 
.
YOU’VE GOT PROBLEMS
Problem 1: Evaluate 
.
The integration formula for inverse tangent is not the only situation in which completing the square will be useful. The formula for inverse sine will also be helpful. You’ll need to be careful with the arithmetic because the radicand in the denominator is 1 – u2 and that nasty subtraction symbol, like a spoiled child, causes a lot of trouble when it isn’t given enough attention.
Example 5: Evaluate 
.
Solution:
1. Complete the square within the radicand while paying attention to how we deal with the subtraction:
2. Do you see why the 36 was added at the end rather than subtracted? The 36 inside the parentheses is affected by the subtraction sign appearing before the parentheses. That means we need to add 36 to keep the balance:
3. Factor 49 from the radicand so that it takes the form 1 – u2:
4. Therefore, 
. Let 
do that 
:
5. Return to the original variable:
Example 6: Evaluate 
.
Solution:
1. Complete the square for 1 – 2x – 2x2: 1 – 2(x2 + x) becomes 
.
2. The radicand is supposed to be of the form 1 – u2, so we’ll need to factor out 
: 
.
3. Now we’ll have to move the 
inside the parentheses:
4. Let’s see what we can do with this. 
becomes 
and this equals 
.
5. Let 
so that 
and 
.
6. The integral transforms to 
, bringing this back to the original variable, 
.
YOU’VE GOT PROBLEMS
Problem 2: Evaluate 
.
Integration by Partial Fractions
Any number of fractional factors could make up the partial fractions. We’ll only consider three cases:
- Nonrepeating linear factors
- Repeating linear factors
- Irreducible quadratic factors
Nonrepeating Linear Factors
Example 7: Evaluate 
.
Solution:
1. As you know, x2 – 9 = (x + 3)(x – 3), so we will write the equation 
.
2. Multiply both sides of the equation by (x + 3)(x – 3) to get 1 = A(x – 3) + B(x + 3).
3. Set x = 3, thus eliminating the first part of the right-hand side of the equation, to get 1 = 6B so 
.
4. Now set x = –3, eliminating the second part of the right side of the equation, to get 1 = –6A or 
.
5. Rewrite 
as 
or 
.
Critical Point
We are looking at cases where the derivative of the denominator is not present in the problem. Given that, when do you use partial fractions versus when do you complete the square? If you can factor the denominator, use partial fractions. If not, go to the process of completing of the square.
Example 8: Evaluate 
.
Solution:
1. Rewrite the denominator as (4x + 1)(x + 3).
2. Solve for the coefficients using partial fractions.
becomes 1 = A(x + 3) + B(4x + 1)
3. Set x = –3 to get 1 = –11B or 
.
4. Set 
to get 
so that 
.
5. We now have 
.
Example 9: Evaluate 
.
Solution:
1. The denominator factors to x(2x + 3)(2x – 3).
2. Create the equation for decomposition:
3. Multiply both sides of the equation by x(2x + 3)(2x – 3):
1 = A(2x + 3)(2x – 3) + Bx(2x – 3) + Cx(2x + 3)
4. Set x = 0: 1 = –9A so 
.
5. Set 
so 
.
6. Set 
so 
.
7. 
becomes 
.
YOU’VE GOT PROBLEMS
Problem 3: Evaluate 
.
Repeated Linear Factors
Integrating 
is straightforward because you can use u-substitution with u = x + 2 and arrive at the answer 
. In this section, we’ll examine rational integrands with three or more factors, with one factor being repeated.
Critical Point
When one or more linear factors are repeated in the denominator of an integrand, it is necessary to include one more fraction on the right side of the equation when decomposing the integrand into partial fractions.
Example 10: Evaluate 
.
Solution:
1. First, take note the x2 is not the derivative, or a multiple of the derivative, of (x + 1)3.
2. When we write the equation to decompose the integrand into partial fractions, we need to realize that x + 1 is a factor three times.
3. Therefore, we’ll need a fraction for x + 1, a fraction for (x + 1)2, and a fraction for (x + 1)3. Each is a linear factor, so the numerators in each case will be a constant.
4. Create the equation for decomposition:
5. Multiply both sides of the equation by the common denominator (x + 1)3:
x2 = A(x + 1)2 + B(x + 1) + C
6. The value of C is easy to determine. Set x = –1: (–1)2 = A(0) + B(0) + C so C = 1.
7. We will not be able to eliminate other coefficients as we try to determine the values of A and B. We will need to create a system of equations.
8. Set x = 1 (because 1 is an easy number): 1 = 4A + 2B + 1 so 4A + 2B = 0.
9. Set x = 2 (another easy number): 22 = 9A + 3B + 1 so 9A + 3B = 3.
10. Solve the system of equations to determine that A = 1 and B = –2.
11. Rewrite the integral into the decomposed fractions (no, they are not zombies):
Example 11: 
.
Solution:
1. The denominator factors to be x(2x + 1)2. The equation to decompose the fraction is:
2. Multiply both sides of the equation by the common denominator to get:
2x + 5 = A(2x + 1)2 + Bx(2x + 1) + Cx
3. Set x = 0: 5 = A.
4. Set 
so that C = –8.
5. Let x = 1 (remember, an easy number): 2(1) + 3 = 5(2(1)+1)2 +B(1)(3) – 8.
7 = 45 + 3B – 8 so that 3B = –30 or B = –10
6. Rewrite the integral into the decomposed fractions:
YOU’VE GOT PROBLEMS
Problem 4: Evaluate 
.
Irreducible Quadratic Factors
The third fractional factor that could make up a partial fraction is quadratics that cannot be factored with the real numbers. Since the numerator of the factors has a degree 1 less than the denominator, the numerators for the quadratic factors will be linear rather than constants.
Example 12: Evaluate 
.
Solution:
1. The denominator factors to x(x2 + 1). The equation to decompose the fraction is:
2. Multiply through by the common denominator to get:
1 = A(x2 + 1) + (Bx + C)(x)
3. Set x = 0: 1 = A + 0 so A = 1.
4. Set x = –1: 1 = 2 + B – C.
5. Set x = 1 (easy number): 1 = 2 + B + C. Solving the system, givens B = –1 and C = 0.
6. Rewrite the integral into the decomposed fractions:
Example 13: Evaluate 
.
Solution:
1. Recall that x3 – 1 = (x – 1)(x2 + x + 1). The equation to decompose the fraction is:
2. Multiply through by the common denominator:
x = A(x2 + x + 1) + (Bx + c)(x – 1)
3. Set x = 1: 1 = A(1 + 1 + 1) + 0 so 
.
4. Set 
so 
.
5. Set 
so 
, and 
.
6. Rewrite the integral into the decomposed fractions:
7. The antiderivative of 
is ln|x–1|. That is not a problem. It’s the second part of the integrand that requires us to be careful.
8. The derivative of x2 + x + 1 is 2x + 1.
9. We don’t have 2x, we have x. Half of 2x + 1 is 
, so we can rewrite 
as 
.
10. The antiderivative of 
is 
.
11. This leaves us with 
.
12. We have a quadratic denominator that cannot be factored with a constant numerator. Time to complete the square in the denominator.
13. Remember, we’re trying to match this denominator to the inverse tangent form 1 + u2. We need to rewrite 
, which equals 
.
14. The antiderivative of 
.
15. Therefore, 
. Wow! That was a mouthful!
YOU’VE GOT PROBLEMS
Problem 5: Evaluate 
.
The Least You Need to Know
- When the denominator of an integrand is of the form ax2 + bx + c, look to see if completing the square is an option you can use to compute the integral.
- Identify quadratic factors in the denominator of the integrand that allow you to rewrite the fraction as the sum and difference of the component fractions.
- You can find the appropriate numerators for the component fractions based on the pattern of the denominator’s factors.