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Krizalid · #3 December 13th, 2007, 10:54 AM

I think liyi is not askin' to solve explicitly the integral. Here's a trick to prove the desired.

Quote:

Originally Posted by liyi

Prove that $\int_0^1\ln\left(1+\sqrt x\right)\,dx=\int_0^1(1-x)\,dx$

$\int_0^1 {\ln \left( {1 + \sqrt x } \right)\,dx} = \int_0^1 {\int_0^{\sqrt x } {\frac{1}{{1 + u}}\,du} \,dx} = \int_0^1 {\underbrace {\int_{u^2 }^1 {dx} }_{\left( {1 - u^2 } \right)}\frac{1}{{1 + u}}\,du} = \int_0^1 {(1 - u)\,du} \,\blacksquare$

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