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## Integral Calculus (2017 edition)

### Course: Integral Calculus (2017 edition) > Unit 6

Lesson 4: Partial fraction expansion# Partial fraction decomposition to evaluate integral

When you are integrating a function in the form of a fraction, it helps to find a way to break apart the expression.

## Want to join the conversation?

- I'm trying to solve some problems on this topic and there's something I don't understand.

Integral of (1/2x+3) = integral of [(1/2)*(1/x+1.5)] = 1/2 integral of (1/x+1.5), right?

But when I try to integrate them, I get different answers:

integral of (1/2x+3) = 1/2*ln |2x+3| + C

1/2*integral of (1/x+1.5) = 1/2*ln |x+1.5| + C

Could you explain what I'm doing wrong? Thanks in advance.(7 votes)- You are not doing anything wrong. Though they don't look it, the two answers are equivalent. The key lies in understanding that C represents ALL constants, not just an unspecified constant. So, C absorbs all of the constants generated by the integration. Thus:

½*ln |2x+3| + C

= ½*ln |2(x+³⁄₂)| + C

= ½ln 2 + ½ln |(x+³⁄₂)| + C

since ½ln 2 is a constant, it is absorbed by C

Thus,

= ½ln |(x+³⁄₂)| + C

So, remember that C is not just some constant that you haven't solved for. It represents the infinitely many solutions generated by every constant that exists. It is only when you get to definite integrals that you can get rid of the C and arrive at a single solution.(31 votes)

- Why can't u just find a and b by choosing an x value that sets one of them to zero?(4 votes)
- You can, but for problems with more complex partial fraction decomposition, specifically irreducible quadratic factors, you won't be able to solve for all the numerators in that way. But for simpler ones I find that to be the easiest way.(10 votes)

- In this video, Sal distributed the A and B into (x-1) and (x+1) respectively, then he factored out the (A+B) from x and used the fact that that equals the coefficient of the x-term and the other expression (B-A) equals the constant term.

In the algebra partial fraction videos, however, Sal would set (x-4) = A(x-1) + B(x+1) then plug in arbitrary values of x so A or B would be multiplied by zero, and he would solve for A and B that way.

My question is this: Which way is better? Or does it really matter which way you go since both ways give the same answer? Personally, I like plugging in x-values and solving that way, but is the way Sal did in this video better than the way I like?

Thanks!(3 votes)- Great question. The method of equating coefficients is technically more correct, but the method of plugging in x-values so that either A or B is multiplied by zero is usually much faster, so I always use it. The reason it is technically wrong is that in order to get (x-4) = A(x-1) + B(x+1), you multiply both sides by (x-1)(x+1), right? But when you do that, you should be very nervous about ever setting x = 1 or x = -1 at any point in the future, because that means that when you multiplied both sides by (x-1)(x+1), you were multiplying both sides by 0, and cancelling 0/0 = 1! That is always a dangerous thing to do and often leads to extraneous or just plain false solutions in many situations. However, as far as the method of partial fractions is concerned, that quick method of multiplying by the denominator and then plugging in "clever" values for x is safe, and you can use it to save time.(5 votes)

- Did anyone else hear the ringtone at4:48:)(4 votes)
- YeahhXD and I replayed the video just to confirm that I wasn't dreaming it up XD(2 votes)

- 5:45why must A+B always be equal to that coefficient and B-A always equal to the constant? I see that for any example I do it's true, but I don't understand why.(3 votes)
- Sorry it's late, just got here. However, remember x+y=x+y, therefore 1x=(A+B)x and -4=(B-A); since your 'x' terms are specified in the equation, hence A+B=1.(1 vote)

- can you do a problem that has a definite integral rather than alway just with the indefinite integral. I mostly have problems that have definite integrals and can't figure out how to take the video from indefinite to definite.(1 vote)
- Is there a shorter way to find the partial fractions? For example, a simple function of the denominators to find the numerators instead of calculating out the whole thing?(2 votes)
- At3:32, he mentions a previous video on partial fraction expansion. Where is that video, or series of videos?(1 vote)
- Hi 20leunge,

I looked up partial fraction decomposition, and found the following as the first video:

https://www.khanacademy.org/math/algebra-home/alg-rational-expr-eq-func/alg-partial-fraction/v/partial-fraction-expansion-1

Hope that helps!(2 votes)

- At8:10, you integrate (5/2)(1/x+1) to 5/2*ln(x+1) explaining that the derivative of the denominator (x+1) is equal to the numerator 1 . Could you please explain this further as I am still very confused?(1 vote)
- For the second part of the new expression of the integral we get at7:50, I distributed 5/2 into 1/(x+1) so that I got 5/(2(x+1))=5/(2x+2). Then I took the integral:

∫5/(2x+2)dx = (5/2)∫2/(2x+2)dx. With u-substitution, I get (5/2)ln(2x+2)+C instead of Sal's (5/2)ln(x+1)+C. Where did I go wrong?(1 vote)- You're not mathematically wrong, though you could argue the (2x+2) bit is not in its simplest form.

ln(2x+2) = ln(2(x+1)), which by the law of logarithms

= ln(2) + ln(x+1)

And, since ln(2) is a constant it can be "rolled up" with C. After multiplying by 5/2, obviously.

In other words the results are the same.(2 votes)

## Video transcript

- [Voiceover] Try to evaluate
the following integral. So assuming you've had a go at it, so let's work through this together. And if at any point you get inspired, always feel free to pause the video and continue on with it on your own. So the first thing that
might have jumped out at you we have a rational expression. The degree in the numerator is the same as the degree in the nominator,
so maybe a little bit of algebraic long division is called for. So let's do that. Let's take X-squared minus one and divide it into X-squared,
put that in a different color, divide it into X-squared
plus X minus five. So X-squared plus X minus five. And so let's look at
the highest degree terms how many times does
X-squared go into X-squared? It goes one time, let me
write this in a new color. Goes one time. One times X-squared minus
one is just going to be X-squared minus one. Now you subtract this
green expression from this mauve expression or I could just add the negative on it, so let me just take the negative of it. So it's going to be
negative X-squared plus one and we're going to get the X-squared's X-square minus X-square is zero, so those cancel out and
we're going to be left with X and the negative five
plus one is negative four. So we have X minus four left over. So we can rewrite the
expression that we're trying to find the antiderivative of. We can rewrite it as one plus X minus four over X-squared minus one. Maybe I'll do that in that purple color since I already used it as the purple. Over X-squared minus one. So we did one thing, now
we have a lower degree in the numerator than we
have in the denominator. And obviously this is
fairly straight forward, take the antiderivative of, but what do we do now? It's not clear if we look
at X-squared minus one it's derivative would be two-X, which is the same degree as this, but it's not X minus four,
so it doesn't look like you u-substitution it's
going to help us with this. So what can we do now? Now we can take out another tool in our algebraic tool kit, we will do partial fraction expansion. Which is essentially
writing this as the sum of two rational expressions that have a lower degree in the denominator. So what do I mean by that? So this term right over here, X minus four over X-squared minus one, we can rewrite that as X minus four over, instead
of X-squared minus one, we can factor this. This is X plus one times X minus one. So let's write that, this is X plus one times X minus one. When we think about
partial fraction expansion we say okay, can we write this as the sum of something, let's call
that A, over X plus one, plus something else, let's call that B, plus B over X minus one. Can we do that? And to attempt to do
that, if we had just add these two things, what would we get? Well we would find a common denominator, which would be X plus
one times X minus one, and so you would have, if
any of this looks unfamiliar I encourage you to review the videos on partial fraction
expansion, because that's exactly what we're doing right over here. But this would be equal to
if you were to add the two your common denominator
would be the product. So it would be X plus
one times X minus one. So the first term I would
multiply the numerator and the denominator times X minus one. So it would be A times X minus one plus B, the second term I'll multiply the numerator and the
denominator times X plus one. And so what do we get? This is going to be equal to AX, maybe I'll do this all in one color. This is going to be equal to AX minus A plus BX plus B and then all of that
over this stuff we keep writing over and over again. Actually, let me just copy and paste this. So copy and paste, I can use
that over and over again. So we have that over that. So let's see, now we
can group the X terms. So we can rewrite this as,
so if we take AX plus BX that's going to be A plus B times X. Then we have a negative A and a B. So plus B minus A, and I'll
just put parenthesis around that just so I kind of group
these constant terms. Then all of that's going to be divided by, good thing I copied and pasted that, X plus one times X minus one. So now this is the crux of
partial fraction expansion. We say, okay we kind of went through this whole exercise on the thesis
that we could do this, that there is some A and
B for which this is true. So if there is some A and
B for which this is true, then A plus B must be the coefficient of the X term right over here. So A plus B must be equal to one, must be equal to this coefficient. And B minus A must be
equal to the constant, must be equal to negative four. Or if they are then we
will found an A and a B, so let's do that. I'll do it up here since I have
a little bit of real estate. A plus B is going to be equal to one, and B minus A or I could write that as negative A plus B is
equal to negative four. We could add the left hand sides and add the right hand sides and then the A's would disappear. We would get two B is
equal to negative three or B is equal to negative three halves. We know that A is equal to one minus B, which would be equal to
one plus three halves, since B is negative three halves, which is equal to five halves. A is equal to five halves. B is equal to negative three halves. And just like that we
can rewrite this whole integral in a way that
is a little bit easier to take the anti or this whole expression so it's easier to integrate. So it's going to be the integral of one plus A over X plus one. A is five halves and so I
could just write that as, let me write it this way, five halves times one over X plus one. I wrote it that way because
it's very straight forward to take the antiderivative of this. Then plus B over X minus one. Which is going to be
negative three halves. So I'll just write it
as minus three halves times one over X minus one. That was this right over here, DX. Notice all I did is I took this expression right over here and I did a little bit of partial fraction expansion into these two, I guess
you could say, expressions or terms right over there. It's fairly straight
forward to integrate this. Antiderivative of one,
it's just going to be X. Antiderivative of five
halves, one over X plus one, is going to be plus five halves, the natural log of the
absolute value of X plus one. We're able to do that
because the derivative of X plus one is just
one, so the derivative is there so that we can
take the antiderivative with respect to X plus one. You could also do
u-substitution like we've done in previous examples, U
is equal to X plus one. And over here, this is going
to be minus three halves times the natural log
of the absolute value of X minus one, by the
same exact logic with how we were able to take the
antiderivative there. And of course we cannot
forget our constant. And there we have it. We've been able to integrate, we were able to evaluate this expression.