Prove that $\displaystyle \int_0^1\ln\left(1+\sqrt x\right)\,dx=\int_0^1(1-x)\,dx$
I think liyi is not askin' to solve explicitly the integral. Here's a trick to prove the desired.
$\displaystyle \int_0^1 {\ln \left( {1 + \sqrt x } \right)\,dx} = \int_0^1 {\int_0^{\sqrt x } {\frac{1}
{{1 + u}}\,du} \,dx} = \int_0^1 {\underbrace {\int_{u^2 }^1 {dx} }_{\left( {1 - u^2 } \right)}\frac{1}
{{1 + u}}\,du} = \int_0^1 {(1 - u)\,du} \,\blacksquare$