Simple integration by parts problem

[revolve]

Homework Statement

<br /> \int \ln(2x+1)dx<br />

Homework Equations

The Attempt at a Solution

u = \ln (2x +1)

du = \frac{2}{2x+1}

dv = dx

v = x<br /> xln(2x+1) - \int \frac {2x}{2x+1}dx <br />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 1/2\int \ln(u)du, with u=2x+1, then integrating ln(u) by parts.

 

[revolve]

Thank you!

So now I have:

<br /> \int \ln(2x+1)dx = x\ln(2x+1) - \int (1 - \frac{1}{2x+1})dx<br />

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:
\int \ln(2x+1)dx = x\ln(2x+1) - \int (1 - \frac{1}{2x+1})dx
=x ln(2x + 1) - x + \int \frac{dx}{2x + 1}
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.