On-line Math 21

5.4  Techniques of integration

5.4.3  Partial fractions

The technique of partial fractions helps solve integrals of nasty rational functions, by splitting up the fraction itself. What we will do first is talk about the basic rational functions we can integrate. Then we'll deal with more complicated expressions, reducing them in terms of these.

The general thing here is to integrate a rational expression:

ó õ P(x) Q(x) dx,

where P(x) and Q(x) are polynomials.

The basic rational functions we can always integrate are of the following two kinds. Note that they have powers of linear terms, or powers of irreducible quadratic terms in the denominator. Note also that the numerator is 1 in the first case, and at most linear in the second. We will reduce all other integrals of rational functions to these types.

1. 
   

   ó õ dx (ax+b)n .

2. 
   

   ó õ (Ax+B)dx ax2+bx+c .

Example 7

ó õ dx 2x+3 .

Here the substitution is u = 2x+3 , so that du = 2dx or dx = du/2 . Then:

ó õ dx 2x+3 = ó õ du/2 u

Example 8

ó õ dx (4x-5)3 .

The second type can cause some trouble. There is a preliminary step, completion of the square, that must be taken for the most complicated ones, and sometimes the problem splits into more than one integral. We'll start with an easy example:

Example 9

ó õ dx x2+3 .

The trick here is to reduce that integral to a standard one, and hopefully

ó õ dx x2+1

comes to mind. The first thing is to turn the 3 into a 1, which you do by simply factoring it out (out of both terms!). Then, substitute as indicated:

ó õ dx x2+3 = 1 3 ó õ dx (x2/3)+1 = 1 3 ó õ dx (x/Ö3)2+1   Now, substitute: u = x/Ö3 = 1 3 ó õ Ö3du u2+1 = 1 Ö3 arctan(u)+C = 1 Ö3 arctan(x/Ö3)+C.

Example 10

ó õ dx 3x2+4

Now a harder one:

Example 11

ó õ (2x+1)dx x2+2x+3 .

The denominator can be reduced to the form u²+a² by completing the square. It then can be dealt with as above. Use the substitution to deal with the linear term in the numerator as well, and split the integral in two. x²+2x+3 = (x+1)²+2 by completing the square, so our first substitution is to set u = x+1 . We will make a second substitution later.

ó õ (2x+1)dx x2+2x+3 = ó õ (2x+1)dx (x+1)2+2 = ó õ (2(u-1)+1)du u2+2 = 1 2 ó õ (2(u-1)+1)du (u/Ö2)2+1 .  Now, set v = u/Ö2 = 1 2 ó õ (2(Ö2v-1)+1)Ö2dv v2+1 = 1 Ö2 ó õ (2Ö2v-1)dv v2+1 = 2 ó õ vdv v2+1 - 1 Ö3 ó õ dv v2+1 = ln(v2+1)- 1 Ö2 arctan(v)+C = ln((u/Ö2)2+1)- 1 Ö2 arctan(u/Ö2)+C = ln(((x+1)/Ö2)2+1)- 1 Ö2 arctan((x+1)/Ö2)+C

There was even a third substitution that I slid in there without mentioning it: t = v²+1 , so dt = 2vdv , which happened to be the numerator of the first integral.

ó õ xdx 2x2+3x+6 .

5.4.4  Real Partial Fractions

In a sense, this isn't even a technique of integration, but a fact from algebra that we use in integration. This technique helps solve integrals of nasty rational functions, by splitting up the fraction itself. There are several parts:

The general thing here is to integrate a rational expression:

ó õ P(x) Q(x) dx,

where P(x) and Q(x)

are polynomials.

Part 1, reduce the fraction.

If deg(P(x)) ³ deg(Q(x)) , the rest of the technique won't work. You gotta divide out. Example:

3x3+4x2+5x+1 x2+1 = 3x+4+ 2x-3 x2+1 .

You should be able to verify that, and recall how to do the long division.

Part 2, General form:

If you have a fraction

P(x) Q(x) ,

with deg(P(x)) < deg(Q(x)) - divide out if you have to - then we set up the general form of the decomposition. For every linear factor, ax+b , in the denominator, put an unknown constant over ax+b on the right. If there is a square, or higher power, of ax+b , then you need terms with denominators ax+b , (ax+b)² , (ax+b)³ ,..., up to the power that (ax+b) has. Each gets a constant numerator. Any irreducible quadratic term, ax²+bx+c with no roots, has a linear term Ax+B as numerator. Repeats get more terms, up to the power occurring, each with a linear numerator.

This is just a recipe for setting up the problem.

7x+3 (x-1)(x+2) = A x-1 + B x+2 7x2+3 (x-1)(x2+2) = A x-1 + Bx+C x2+2 7x2+3 (x-1)2(x+2) = A x-1 + B (x-1)2 + C x+2 7x2+3 (x-1)(x2+2)2 = A x-1 + Bx+C x2+2 + Dx+E (x2+2)2 .

The deal is, these always have solutions. You just follow the recipe, like in a cookbook. There is a reason why that will always work, but we will keep that from you - it's best explained in a course on complex analysis.

Part 3. Find the unknown numbers.

This is an exercise in algebraic manipulation. Let's look at the second one. The others are similar - and maybe the first one is too easy. If

7x2+3 (x-1)(x2+2) = A x-1 + Bx+C x2+2 ,

what are the numbers A , B , and C ? You find them by putting the right-hand side back under the common denominator:

7x2+3 (x-1)(x2+2) = A x-1 + Bx+C x2+2 = A(x2+2) (x-1)(x2+2) + (Bx+C)(x-1) (x2+2)(x-1) = A(x2+2)+(Bx+C)(x-1) (x2+2)(x-1) ,

then you compare numerators. The numerators gotta be the same, since the denominators are. And, they gotta be the same for all x . So,

7x2+3 = A(x2+2)+(Bx+C)(x-1).

The easiest way to solve this is to say that the two sides are the same for all x , so they are at the x 's that are easy. Like, when x = 1 , you get: 10 = A3

since the (x-1) becomes 0 . So, A = 10/3 . One down.

Then, try - hmm, no more really easy ones. Well, x = 0 is still not too bad, then:

3 = A(2)+(C)(-1) = 20/3-C.

So, C = 11/3 . Finally, take x = -1 - it keeps the arithmetic simple. That gives:

10 = 7+3 = A(1+2)+(-B+C)(-1-1) = (10/3)(3)+(-B+11/3)(-2),

so B = 11/3 .

Some people get concerned when we set x = 1 in this problem, since that is a zero of the denominator. However, the point is that the numerators, after you put the right-hand side back under the common denominator, must be equal as polynomials, so they are the same even at x = 1 .

Part 4 Actually solve the integral.

This will usually be the easy part. Let's try it for the example above.

ó õ 7x2+3 (x-1)(x2+2) dx = 10 3 ó õ dx x-1 + 11 3 ó õ x x2+2 dx+ 11 3 ó õ 1 x2+2 dx = 10 3 ln(x-1)+ 11 6 ln(x2+2)+ 11Ö2 6 arctan(Ö2x)+C,

by the techniques we went over earlier.

Copyright (c) 2000 by David L. Johnson.