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$Old$ December 13th, 2007, 08:53 AM

liyi $liyi is offline$

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$Default$ Integration proof

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

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#2

$Old$ December 13th, 2007, 09:52 AM

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

$t = 1+\sqrt{x}$

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#3

$Old$ December 13th, 2007, 10:54 AM

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

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Originally Posted by liyi $View Post$

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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