Originally Posted by

**galactus** Here is an integral from the site that isomorphism posted. I thought I would try my hand at this one using Kriz's double integration technique.

$\displaystyle \int_{0}^{1}\frac{ln(x)}{1+x}dx=\frac{-{\pi}^{2}}{12}$

We know that $\displaystyle ln(x)=\int_{1}^{x}\frac{1}{u}du$

Then we get:

$\displaystyle \int_{0}^{1}\int_{1}^{x}\frac{1}{u(x+1)}dudx$

Now, here is where I am hung up on switching the limits of integration, coming up with the new limits, and thus coming up with a nice workable integral.

It looks similar to $\displaystyle \int_{0}^{1}\frac{ln(x+1)}{x^{2}+1}dx=\frac{\pi}{8 }ln(2)$