# Thread: evaluate the integral

1. ## evaluate the integral

evaluate the integral from 0 to 1/2 of (ln(2x+1))/(4x2+1)dx. any help is greatly appreciated, thank you in advance

2. ## Re: evaluate the integral

Did you mean $\int_0^{1/2}\ln \frac{2x+1}{4x^2+1}dx$?

If so, remember log(a/b)=log a - log b.

You can then use integration by parts on ln (4x^2+1).

3. ## Re: evaluate the integral

no, I edited it appropriately as to eliminate any confusion. btw, how do you create an integral within the text?

4. ## Re: evaluate the integral

Originally Posted by tannergilligan
no, I edited it appropriately as to eliminate any confusion.
In that case, take a look at integrate log&#40;2x&#43;1&#41;&#47;&#40;4x&#94;2&#43;1&#41; dx - Wolfram|Alpha .

Originally Posted by tannergilligan
btw, how do you create an integral within the text?
LaTeX Help

5. ## Re: evaluate the integral

I have already used the integral calculator on wolfram, but that really isn't a solution. I know in this thread: <http://mathhelpforum.com/calculus/42341-integrate-ln-tanx-1-a.html> , he says on the 3rd to last line that someone had already solved this integral earlier, but I have no idea where this solution is. any help is greatly appreciated.

6. ## Re: evaluate the integral

If it makes any difference to you guys, it can also be written as (1/2)[integral from 0 to 1 (ln(u+1))/(u^2+1)du] using the substitution u=2x. not a big difference, but enough to make it simpler to look at.