By this point, we can integrate functions until the cows come home. We were stopped a little bit by the difference of squares, but we figured it out. But we have one more major obstacle in our road to integration freedom - funny-looking rational functions.
We know how to find common denominators to add or subtract fractions. If we had 2/(x-1) - 1/(x +2), it would be no trouble to multiply both of them by what they would need to get their common denominator and subtract your little heart out. Therefore, if we saw a fraction with a denominator that could be factored, we could reasonably assume that we could split it apart.
Assume we have two functions - P(x) and Q(x), and both of them are polynomials. If we're trying to take P(x)/Q(x), then, as long as that fraction is proper (P is a lower degree than Q), then we can write this fraction as the sum of two simpler fractions. Once we have the sum of two easy fractions, then we have a 1/u and we can integrate without much fuss. If P is a higher degree than Q, then we just do a bit of long division first and voila. We'll all just pretend that you've been practicing your long division of polynomials since you got out algebra 2. You know, since you have to do that a lot.
So, when we're dividing up into partial fractions, there are two basic forms it will take:
either A/(ax+b)ⁿ
or (Ax+B)ⁿ/(ax²+bx+c)ⁿ
You should be able to divide your denominator into terms that fit that form. This leads to a few different scenarios.
SCENARIO 1 - The denominator consists of distinct linear factors.
It's possible that when you factor out your denominator, you'll end up with linear terms (Like ax+b) and none of them repeat. For example, 2x²+17x+21 factors out into (2x+3)(x+7).
If this is the case, then split it up into the sum of different constants with each of those factors as the denominator. Our above case would give us A/(2x+3) + B/(x+7) where A and B are two different constants (Though they may turn out to be equal. You never know in the crazy world of math. Woo!).
After that, leaving your A and B as variables, add those fractions back together with a common denominator. Let's say we were factoring (x+3)/(2x²+17x+21) so we can use our above factoring.
Once we add the fractions back together, if we keep our original function on the left side of our equation, we get:
(x+3)/(2x²+17x+21) = Ax+7A+2Bx+3B/(2x²+17x+21). (after multiplying to get common denominators.)
Since the two sides of equation are equal, and their denominators are equal, then their numerators have to be equal. Since no more work needs to be done on the denominators to make them equal, we'll ignore them for now and work on our numerators. After taking out the denominator, we're left with:
x+3 = Ax+7A+2Bx+3B
Since A and B are only constants, we know that when we add up all of the coefficiants of x, they'll give us 1, and all of the leftover constants will add up to 3. So let's separate our constants by the ones that are going to pertain to our x and our 3.
x +3 = x(A+2B) + (7A+3B)
Now we know that A+ 2B = 1 and 7A+3B = 3.
All that's left to do is solve our little system of equations, plug in the A and B values we find back into our equation up top, and we know have much tidier and easier to integrate fractions.
That's the most straightforward case, and the following cases are just variations on the above.
SCENARIO 2, ELECTRIC BOOGALOO- Linear factors, but some repeat.
So let's say your denominator factors, but some of them repeat themselves. If your denominator was x²+4x+4, then it factors out into (x+2)².
At first, you would think you would do the same as the above and have A/(x+2) + B/(x+2), but that's not the case. Instead, you list each "incarnation" of the term that can be factored out of it. It's hard to say in words, but this one would become:
A/(x+2) + B/(x+2)²
If the denominator was (x+2)^7, it would look like:
A/(x+2) + B/(x+2)² + C/(x+2)³... G/(x+2)^7.
So you make a factor out of each "Step" on the way to the power of the denominator. It's a little weird, and the book decides not to explain why because I'm sure way back in algebra that was explained.
After that, you proceed as normal. Hopefully no workbook is ever sadistic enough to give you one that's 7th degree, because that would be rather tedious.
SCENARIO 3D - Irreducible Quadratic Factors (dunnnn)
The above are what to do once you've got it reduced to linear factors. But, if fate is not looking kindly on you, you may find your denominator still quadratic, but irreducible. If this is your destiny, you proceed on, but instead of just an A as your numerator, it will instead be Ax+B/(whatever evil quadratic thingy).
Then, again, you proceed as normal. Though usually on these you'll have some less than pretty denominators still, usually the sum of two squares, but there is a formula for that and it is probably in the back of whatever calculus book you're using.
SCENARIO 4: Shin Megami Tensei - Repeated irreducible quadratic factors
Ugh....math words...make me tired. We need a math vocabulary overhaul.
Anyway.
If these IQFs repeat, you treat it the same as when linear factors repeat, listing each step along the way to whatever power it is.
These typically take up entire sheets of paper and sometimes two lines to write out, so you probably encounter too many of these. But, if you do, have patience, because mis-writing something is where you're most likely to make a mistake here after copying down a number for the 100th time.
And there we go. We now have the complete basic arsenal of integration tools. The next couple of sections are just strategies for deciding how to go about solving a given integral problem, how to use the tables in the back for particularly ugly integrals, and how to use computer programs like Maple and Mathematica that are far too expensive for me, so I'm going to skip over them since there's not a whole of content there.
That, and I have a test on this chapter in a few days and I need to get it over with, so....yeah.
Peace.
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