- #1

- 11

- 0

*This has been posted in another, but the wrong, section of the forum. That's why there are already quotes in it.*

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 ;)From ordinary division … (2x + 1)/2 = x + 1/2, so x - 1 = (2x + 1)/2 - 3/2

But isn't it supposed to becomeFrom the chain rule

*∫1/(2x+1)dx*? Since the dirative of

*ln(x) = 1/x*?

## 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

= -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