# Homework Help: Evaluate the integral, integration by parts

1. Aug 19, 2008

### afcwestwarrior

1. The problem statement, all variables and given/known data
∫ln (2x+1) dx

2. Relevant equations
∫u dv=uv- ∫ v du
integration by parts

3. The attempt at a solution
u= ln (2x+1) dv=dx
du=???? v=x

ok did i choose the right u and how do i derive it, do i have to use the chain rule

π²³ ∞ 0° ~ µ ∑ Ω √ ∫ ≤ ≥ ± # … θ φ ψ ω Ω α β γ δ ∂ ∆ ε λ Λ Γ ô

2. Aug 19, 2008

### Dick

Yes. Use the chain rule.

3. Aug 19, 2008

### nicksauce

You did choose the 'right' u, although a substitution like g=2x+1 might make things easier, though it's not necessary. To find du, just apply the chain rule, and use the fact that d/dx(ln(x)) = 1/x.

4. Aug 19, 2008

### afcwestwarrior

ok, i'll try it out

5. Aug 19, 2008

### afcwestwarrior

so i used the chain rule and i did this
ln (2x+1)= 1/x * (2x+1) * 2= du

did i do it correctly and du is 4x+4/x

6. Aug 19, 2008

### Feldoh

If $$u = ln(2x+1)$$ then $$du = \frac{2 dx}{2x+1}$$

Last edited: Aug 19, 2008
7. Aug 19, 2008

### afcwestwarrior

so i did it wrong

8. Aug 19, 2008

### afcwestwarrior

i don't get how you got 2/2x+1

9. Aug 19, 2008

### Feldoh

Looks like it :-/

It might make sense to do the derivative like this:

First: g(x)=2x+1 and f(x)=ln(g(x)) so then the chain rule is:

$$\frac{d}{dx}f(x) = g'(x)*f'(g(x))$$

g'(x) = 2 and f'(g(x)) = 1/g(x) = 1/(2x+1)

From there is should be pretty easy to follow.

10. Aug 19, 2008

### afcwestwarrior

oh ok, i didn't know i could've used g for 2x+1 that's what the other guy was telling me, i'm really rusty on my calculus, thanks

11. Aug 19, 2008

### afcwestwarrior

this is what i got so far
u= ln (2x+1) dv=dx
du= 2/2x+1 v=x

ln (2x+1) (x) - ∫x (2/2x+1

now i'm integrating ∫x (2/2x+1 dx
u=x dv=2/2x+1
du=dx v=?????
I'm stuck here how do i antiderive 2/2x+1

Last edited: Aug 19, 2008
12. Aug 19, 2008

### afcwestwarrior

is it 1/2 (2x+1)

13. Aug 19, 2008

### afcwestwarrior

ok i've decided to use u=2/2x+1 dv=x
du=2x-4/(2x+1)^2 v=dx

14. Aug 19, 2008

### afcwestwarrior

i forgot about factoring do i have to factor du now

15. Aug 19, 2008

### Dick

If you want to integrate (2x)/(2x+1) don't use parts. Use the substitution u=2x+1 again. So 2x=u-1. Can you put it all together?

16. Aug 19, 2008

### afcwestwarrior

ok i'll give it a shot

17. Aug 19, 2008

### afcwestwarrior

ok what do i do after

18. Aug 19, 2008

### Dick

After what? What you done so far? I'm just suggesting you do it as a u-substitution. You've done that before, right? What's the integral in terms of u?

19. Aug 19, 2008

### afcwestwarrior

du is the term

20. Aug 19, 2008

### Dick

??? You have 2x/(2x+1)dx. You want to replace all of the parts with their equivalents in terms of u. What's the new u integral?