Now we are in to a section of integration that many people enjoy doing since it does not really require a lot of messy calulus work. It involves a lot of algebra (but I'm sure you are a pro at that already!). It is the method of integration by partial fractions. This method involves a ratio of polynomials as seen below:
where P and Q are polynomials. Again, if the degree of P is larger than the degree of Q, we are to use long divison to get it into a form we can work with (though working with partial fractions, this will not happen that frequently).
When expanding the terms for the partial fractions, we must know that the corresponding numerator must be less than the denominator. To show this, we only have two cases for partial fractions:
For this technique, there are two cases that can occur:
CASE 1: Linear factors in the denominator.
For each term in the denominator of the form (ax + b)n there will be n partial fractions. An example of this will be the following:
Notice that each of the terms in the denominator are linear factors so the numerators in the expansion will be constant terms (A, B, C ...). And also notice that the (x+2) term is repeated twice. This is because the exponent in the original denominator tells you how many times that term will appear in the expansion. It always starts from its lowest power and proceeds to its highest when expanding the terms. So if the power was 4, there will be (x+2), (x+2)2 , (x+2)3 , (x+2)4 and so on.
More generally, the partial fraction expansion can look like:
CASE 2: Quadratic factors in the denominator.
For each term in the denominator that is of the form (ax2 + bx + c)m there will be m partial fractions. An example of this is the following:
Similar to case 1, you are making sure that each term in the expansion has a numerator that is less than the denominator and also that each term in the original denominator is represented the correct number of times. Observe in the above expansion that the final two terms of the expansion have a linear numerator because their denominator is quadratic.
More generally, the partial fraction expansion can look like:
Example 1
Easily enough, the denomiator is already factored so this is an example of case 1 above. We find the partial fraction expansion for the term in the integral to be:
But now we need to solve for A, B and C. We cannot simply leave it like that as a final answer. So the technique is to multiply by the LCD of the original fraction on both sides to get a term where you can plug in values to solve for A,B and C. Upon multiplying it out, we get the following term:
Now, there are three ways of solving this equation above; the first is to plug in appropriate values of x to eliminate certain terms to solve for A,B and C; the second way is to use a system of equations and the third way is to perhaps use a combination of both (if you have a lot of terms to solve for, the third way is quite useful). For the purpose of this example, we will use the first method which is to plug in appropriate values of x.
And we find that C must equal 3.
And we find that A must equal 1.
And we conclude that B must equal -1. So now, the known values of A,B and C are A = 1, B = -1, C = 3. Now, the integral will become:
We can rewrite the integral on the right as the integral term by term:
Now, evaluate the right integrals and solve for your final answer of:
Example 2
First, we notice that the denominator is not in factored form so we must factor the denominator to get the following integral:
The partial fraction expansion will produce the following:
Multiply by the LCD on both sides to get the equation:
Be sure to multiply out on the right side of the equation and group the like terms to get:
We can now solve this by the systems of equations:
It is immediately obvious that B will equal -2 (the last equation). Now take that and plug into the second equation above:
And we see that A will equal -2 as well. Now take that value of A and place it in the first equation above:
So we see that C will equal 2 to make that equation equal to 0. The values for A, B and C are: A = -2, B = -2, C = 2. Now the integral will become:
Let's take a moment to observe the first integral on the right side above. Here is a way to solve this integral: you may split the integral into the following:
This is perfectly legal as it is seen in algebra when dealing with fractions and LCDs. Now piece these parts together with the other piece on the right to get:
Solve each integral to get a final answer of:
Example 3
This is an integral of case 2 above with the quadratic term in the denominator unfactorable. We get the partial fraction expansion of:
Next, we get the following equation to work with:
The below is optional because the A,B and C can be found without this additional expansion:
Either way, we can solve this by choosing appropriate values for x in the equation.
We see that A must be 1.
We can see that C must be 3.
We can see that B must be 2. The values for A, B and C are: A = 1, B = 2, C = 3. Now plug in to the integral and solve accordingly.
The leftmost integral is simply an ln. Observing the rightmost integral, we can see that it is made up of an ln but not quite so obviously. It requires some manipulation. Observe the manipulation below:
We can now split the integral as we did in the previous example to get:
The first of these is an ln because the derivative of the bottom is the numerator (2x + 2) and the right integral above needs to be done by completing the square (yes, none of the previously learned techniques go away!!!). Here is that integral:
Putting all the pieces together, we get the following answer:
You can see that this problem consisted of numerous techniques and manipulations and was a more difficult problem. Doing more integrals by partial fractions will certainly help you keep sharp on the previously learned techniques.