In this tutorial we are going to learn about Partial Fractions. Partial fractions is a technique used to solve integrals but it is really grounded in algebra. So, we are going to be doing a lot of algebra and not that much integration. The point of partial fractions is to solve integrals where we have something like one polynomial, or maybe just a constant, divided by another polynomial. What we are going to do is factor the denominator polynomial and then we will do some algebra to separate this fraction into other pieces, into simpler fractions that are supposed to be easier to integrate.
You can always factor the
denominator down into linear (degree one) and quadratic (degree two)
factors. If you have factors that are cubic or higher, you should be
able to factor more.
To factor a cubic, test factors of
the constant term divided by factors of the leading coefficient as
roots. A quick way to check if they work is synthetic substitution
(also known as synthetic division).
Sometimes you can figure out the A
and B by plugging in a particular value of x to
both sides of the equation. (Choose a value that makes one of the
expressions equal to 0.)
If the degree of the numerator is
the same or greater than the degree of the denominator, use long
division first. Based on the results of your long division, split
your integral into a polynomial part (easy to integrate) and a
rational function that you can apply partial fractions to.
If you have a quadratic in the
denominator that doesnt factor, then follow the following steps
= denominator first.
Then separate into
On the first part, use ln |u|.
On the second, complete the
square and use a trigonometric substitution. It should always be a
tangent substitution. (If it is a sine or secant substitution, then
you could have factored the quadratic.)
Expand [4/((x + 1)(x + 2))] using Partial Fractions
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
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We can solve the first one by doing u=x+2 dU=dX, so we get the integral of dU/u, and that is equal to ln(abs(u)) which is the ln(abs((x+2))).0319
The second one is solved in exactly the same technique, so I will skip over the details of that.0342
This minus is from the negative sign above the ln(x+3) + a constant.0353
The integration there at the end was very easy. 0364
The important part that we are trying to learn here is the algebra, where you start with a polynomial in the denominator, you factor it out and then you try to split this expression up into two parts with an a and a b.0370
Then, you combine those two parts back together, that gives you two equations for a and b.0386
You solve those two equations for a and b, you substitute them back into the original expression, and you get an integral that hopefully turns out to be easy to solve.0397
Let us try another example that gets a little more complicated.0407
Example 2 here, again we have a polynomial over a polynomial. 0411
We can write (2x2 - 7x)/(x2-6x+8) as our answer 2 plus a remainder term (5x-16)/(x2-6x+8).0527
Now, the important thing here is that the degree of the numerator is now 1, the degree of the denominator is now 2.0551
The denominator now has a bigger degree and partial fractions is going to work.0566
Remember, partial fractions does not work if the degree of the numerator is bigger than the degree of the denominator.0569
That is no longer true after we do polynomial division and we can expect the partial fractions to work.0580
I am not going to worry about the 2 for a while, because I know that 2 is going to be easy to integrate.0584
I am going to do partial fractions now on 5x-16 over that denominator, and I can factor that into (x-2) × (x-4).0598
I am going to try to separate that out into a/(x-2) + b/(x-4).0614
Just like we did before, we are going to end up with a × (x-4) + b × (x-2), that comes from trying to combine these terms over a common denominator, is equal to the numerator on the other side (5x-16).0626
The degree is 1 in the numerator, the degree of the denominator, the power of the highest term, is 2.0824
So that is ok, the numerator has a smaller degree than the denominator so we expect partial fractions to work.0831
We are going to go ahead and try to factor the denominator.0843
(x2-6x+9), that factors as (x-3)2 and if you are going to try to separate that out using partial fractions, your first instinct might be to try a/(x-3) + b/(x-3) the same way we did the previous examples.0845
The secret to fixing this is to have a/(x-3) but b/(x-3)20914
That is kind of the secret to this problem, is to put in that square on (x-3).0931
Then if you try to find, if you try to match up the numerators, we get (3x-7)=a × (x-3).0935
To get the common denominator, and then + b because if you wanted to combine those two, our common denominator would be (x-3)2 and b is already over that denominator, so we would not have to multiply b by anything to get that common denominator. 0948
Then, we get ax-3a + b = (3x-7), so ax plus, let me separate out terms (3a + b)=3x-7 and again you get two equations and two unknowns except this one is very easy, you get a=3 and minus 3a+b=-7.0970
Again, solve that using anything you want from high school algebra and of course you get a=3 and you plug that in, it works out to b=2.1003
So, our integral turns into the integral of a/(x-3) + , now b=2/(x-3)2, all of that times dX.1021
Now, we have learned already how to integrate 3/(x-3), that is ln(abs(x-3)).1045