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### Course: Class 12 math (India)>Unit 9

Lesson 8: 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.
• 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.
• Why can't u just find a and b by choosing an x value that sets one of them to zero?
• 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.
• 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!
• 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.
• Did anyone else hear the ringtone at :)
• YeahhXD and I replayed the video just to confirm that I wasn't dreaming it up XD
• why 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.
• 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?
• At , he mentions a previous video on partial fraction expansion. Where is that video, or series of videos?
(1 vote)
• At , 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 at , 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.