- #1
SYoung
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This has been posted in another, but the wrong, section of the forum. That's why there are already quotes in it.
Hello,
I 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 advance,
Young
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
I'm still not sure about the 'ordinary division'.
Thanks again,
Young
Hello,
Homework Statement
I 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 advance,
Young
Homework Equations
Wow that has been a long 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 ordinary division … (2x + 1)/2 = x + 1/2, so x - 1 = (2x + 1)/2 - 3/2
But isn't it supposed to become ∫1/(2x+1)dx? Since the dirative of ln(x) = 1/x?tiny-tim said:From the chain rule
The Attempt at a Solution
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
I'm still not sure about the 'ordinary division'.
Thanks again,
Young