OK i will post my solution since nobody else replied

$\displaystyle I = \int\limits_0^{\frac {\pi }{4}} {\ln \left( {1 + \tan x} \right)dx}$

$\displaystyle I = \int\limits_0^{\frac {\pi }{4}} {\ln \left( {1 + \tan \left(\frac{\pi}4 - x\right)} \right)dx} $

$\displaystyle I =\int\limits_0^{\frac {\pi }{4}} \ln \frac2{1 + \tan x} dx$

$\displaystyle I = \int\limits_0^{\frac {\pi }{4}}\ln 2 \,dx - I$

So $\displaystyle I = \frac{\pi}8 \ln2$

**Part 2:**
$\displaystyle \int^1_0\frac{x^4(1-x)^4}{1+x^2} dx$

This is a funny integral... not hard... but the answer is interesting

People who solve it can comment on the answer rather than showing the answer