Simple integration by parts problem

[revolve]

Homework Statement

[tex]
\int \ln(2x+1)dx
[/tex]

Homework Equations

The Attempt at a Solution

[tex]u = \ln (2x +1)[/tex]

[tex]du = \frac{2}{2x+1}[/tex]

[tex]dv = dx[/tex]

[tex]v = x[/tex][tex]
xln(2x+1) - \int \frac {2x}{2x+1}dx
[/tex]I'm not sure how to proceed. Do I separate the fraction in the integrand or do long division?
I think I separate the fraction but I don't remember how. So I guess the question is how do I separate the fraction in the integrand?

 

[ideasrule]

You can rewrite the numerator as 2x+1-1, then separate the fraction into 1-1/(2x+1). A slightly easier way of doing this problem would be to rewrite the original integral as [tex]1/2\int \ln(u)du[/tex], with u=2x+1, then integrating ln(u) by parts.

 

[revolve]

Thank you!

So now I have:

[tex]
\int \ln(2x+1)dx = x\ln(2x+1) - \int (1 - \frac{1}{2x+1})dx
[/tex]

I'm stuck on the next step. I think I need to remove the 1 somehow from in front of the fraction.

 

[Mark44]

Continuing where you left off:
[tex]\int \ln(2x+1)dx = x\ln(2x+1) - \int (1 - \frac{1}{2x+1})dx[/tex]
[tex]=x ln(2x + 1) - x + \int \frac{dx}{2x + 1}[/tex]
The -x term came from integrating -1 with respect to x. The last integral can be done with a simple substitution, u = 2x + 1, du = 2dx.

Don't forget the constant of integration.

 

[revolve]

Perfect! Thank you.