Integrals with partial fractions

In summary: I'm bad at it I hope...In summary, the conversation was about understanding a step in the integration of a given function. The expert provided a brief explanation of how to divide and use the chain rule to obtain the desired result. The learner also asked for a similar example to clarify the process.
  • #1
SYoung
11
0
Hello,


I'm don't understand a step in the following integral:

∫(x-1)/(2x+1)dx = ∫(1/2)dx − (3/2)∫1/(2x+1)dx = (1/2)x − (3/4)ln|2x+1| + C

The first step, where you get the 2 integrals ∫(1/2)dx and -(3/2)∫1/(2x+1)dx
Where do (1/2)dx and -(3/2) come from?
And where does (3/4) come from in the last part?

My best guess is that it's done with partial fractions. But even so, I have no clue how.


Thanks in forward,
Young
 
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  • #2
Welcome to PF!

Hello Young! Welcome to PF! :smile:

(btw, it's "thanks in advance" :wink:)
SYoung said:
Where do (1/2)dx and -(3/2) come from?

From ordinary division … (2x + 1)/2 = x + 1/2, so x - 1 = (2x + 1)/2 - 3/2
And where does (3/4) come from in the last part?

From the chain rule :smile:
 
  • #3


EDIT: I think I posted this in the wrong section of the forum. My apologies.
So I have no need for further instructions in this thread. Thank you.

tiny-tim said:
Hello Young! Welcome to PF! :smile:
Thanks! :smile:

tiny-tim said:
From ordinary division … (2x + 1)/2 = x + 1/2, so x - 1 = (2x + 1)/2 - 3/2
Wow that has been a looooong time! Could you please show it step by step how it's done in this case? I hope I can remember the whole thing if I see it ;)

tiny-tim said:
From the chain rule :smile:
But isn't it supposed to become ∫1/(2x+1)dx? Since the dirative of ln(x) = 1/x?

EDIT:
nvm the chainrule, figured it out ;)
∫ -3/(2 (1+2 x)) dx
= -3/2 ∫ 1/(2 x+1) dx
For the integrand 1/(2 x+1), substitute u = 2 x+1 and du = 2 dx:
= -3/4 ∫ 1/u du
The ∫ of 1/u is ln(u):
= -(3 ln(u))/4+C
Substitute back for u = 2 x+1:
= -3/4 ln(2 x+1)+C


Thanks again,
Young
 
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  • #4


SYoung said:
EDIT: I think I posted this in the wrong section of the forum. My apologies.
From your other posts, it looks like you have found the right subforum for calculus homework. So no problem. I'll move this thread to the right place too, so we can set a good example for others :smile:

Wow that has been a looooong time! Could you please show it step by step how it's done in this case? I hope I can remember the whole thing if I see it ;)
Well, not to do your work for you, I'll show a similar example.

Expand (2x-3)/(6x+5) into partial fractions.

While we could do long division here, I'll follow what tiny-tim seemed to be doing...

Approach: we need to get an equation with (2x-3) on one side and (6x+5) on the other side ... then we can divide by (2x-3) to come up with an expression for their ratio.

Starting with 6x+5, we divide by 3 in order to get "2x", the leading term in 2x-3:

(6x+5)/3 = 2x + 5/3​

To make the left-hand-side into the desired 2x-3, we must next subtract 14/3 from both sides:

(6x+5)/3 - 14/3 = 2x-3​

From there you can get to

(2x-3)/(6x+5) = ... ?​
 
  • #5


Redbelly98 said:
Approach: we need to get an equation with (2x-3) on one side and (6x+5) on the other side ... then we can divide by (2x-3) to come up with an expression for their ratio.

Starting with 6x+5, we divide by 3 in order to get "2x", the leading term in 2x-3:

(6x+5)/3 = 2x + 5/3​

To make the left-hand-side into the desired 2x-3, we must next subtract 14/3 from both sides:

(6x+5)/3 - 14/3 = 2x-3​

From there you can get to

(2x-3)/(6x+5) = ... ?​


Sure any example is fine!

Ah so if you fill in ((6x+5)/3 - 14/3) in stead of (2x-3) you get:

((6x+5)/3 - 14/3)/(6x+5) = 1/3 - 14/3


So what you basicly want is to have the same terms (or do you call it values?) in the denominator and numerator, right? Allowing you to easily divide it


Funny, still having trouble with divisions while I use integrals all day long because of my study... heh oh well nobody is perfect:rolleyes:
I'm sorry if I understood you wrong. Daily english is somewhat different from all these mathematical terms ^^ English isn't my first language
 

FAQ: Integrals with partial fractions

What are integrals with partial fractions?

Integrals with partial fractions are a method of simplifying and solving integrals of rational functions by breaking them down into smaller, simpler fractions.

When are integrals with partial fractions used?

Integrals with partial fractions are commonly used when the integrand is a rational function (a polynomial divided by another polynomial), as they can simplify the integration process.

How do you find the partial fraction decomposition of a rational function?

To find the partial fraction decomposition, the denominator of the rational function must first be factored into linear factors. Then, the coefficients of each linear factor are determined using a system of equations and the method of undetermined coefficients.

What are the different types of partial fractions?

The two types of partial fractions are proper fractions, where the degree of the numerator is less than the degree of the denominator, and improper fractions, where the degree of the numerator is equal to or greater than the degree of the denominator.

Are there any special cases when using integrals with partial fractions?

Yes, there are two special cases to be aware of when using integrals with partial fractions: repeated linear factors and non-repeating irreducible quadratic factors. These cases require additional steps in the decomposition process.

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