Evaluate the integral, integration by parts

[afcwestwarrior]

Homework Statement

∫ln (2x+1) dx

Homework Equations

∫u dv=uv- ∫ v du
integration by parts

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

 

[Dick]

Yes. Use the chain rule.

 

[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.

 

[afcwestwarrior]

ok, i'll try it out

 

[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

 

[Feldoh]

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

 

[afcwestwarrior]

so i did it wrong

 

[afcwestwarrior]

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

 

[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.

 

[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

 

[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

 

[afcwestwarrior]

is it 1/2 (2x+1)

 

[afcwestwarrior]

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

 

[afcwestwarrior]

i forgot about factoring do i have to factor du now

 

[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?

 

[afcwestwarrior]

ok i'll give it a shot

 

[afcwestwarrior]

ok what do i do after

 

[Dick]

ok what do i do after

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?

 

[afcwestwarrior]

du is the term

 

[Dick]

du is the term

? 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?

 

[afcwestwarrior]

i don't know I'm confused

 

[afcwestwarrior]

is this what u mean u-1/2x+1

 

[afcwestwarrior]

or u-1/u

 

[Dick]

2x+1=u, right? 2x=u-1, right? du=2dx, so dx=du/2, right? That makes it (u-1)/u*du/2. Now (u-1)/u=1-1/u, yes? Can you integrate (1/2)*(1-1/u)du?

 

[Dick]

or u-1/u

Right. Use parentheses tho. You mean (u-1)/u. Not u-(1/u). Don't forget to substitute for the dx part.

 

[afcwestwarrior]

integrate by parts
1/2* x - ∫x *2dx

 

[Dick]

integrate by parts
1/2* x - ∫x *2dx

You switched problems? What are you doing?

 

[afcwestwarrior]

woops, man i took calculus last year and now I'm taking it again and i forgot everything

 

[Dick]

woops, man i took calculus last year and now I'm taking it again and i forgot everything

I couldn't agree more strongly. We are at 28 posts on a not very hard problem. Jeez. Last I recall, we were trying to figure out the integral of (1/2)*(1-1/u)du. Can you help me with that?