Integrals with partial fractions

[SYoung]

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

 

[tiny-tim]

Welcome to PF!

Hello Young! Welcome to PF!

(btw, it's "thanks in advance" )

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

 

[SYoung]

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.

Hello Young! Welcome to PF!

Thanks!

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 ;)

From the chain rule

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

 

[Redbelly98]

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

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) = ... ?​

 

[SYoung]

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 basically 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
I'm sorry if I understood you wrong. Daily english is somewhat different from all these mathematical terms ^^ English isn't my first language