Integrate $\ln(1 + x^\frac{1}{2})$ from $0$ to $1$ This integral is from an Integration Contest. The substitution $x^2 = u$ allowed me to evaluate the integral but I keep getting $\frac{1}{2} - \ln2$ as the answer as opposed to just $\frac{1}{2}$ which was the posted answer. Any help would be appreciated. 
https://www.wolframalpha.com/input/?i=integrate+ln(1+%2B+rootx)+from+0+to+1
Edit: Can some provide a solution?
The substitution x^2 = u trivializes the integral into 2u(1+u), integrating by parts allows me to evaluate the indefinite integral, but I seem to keep messing up plugging the bounds.
 A: you are  in right way,by partial integration we get $$\int _{ 0 }^{ 1 }{ \ln { \left( 1+\sqrt { x }  \right)  } dx } ={ x\ln { \left( 1+\sqrt { x }  \right)  }  }_{ 0 }^{ 1 }-\int _{ 0 }^{ 1 }{ x\frac { \frac { 1 }{ 2\sqrt { x }  }  }{ 1+\sqrt { x }  }  } dx=\\ =\ln { 2 } -\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { \sqrt { x }  }{ 1+\sqrt { x }  }  } dx=\ln { 2 } -...\\ \sqrt { x } =t\\ x={ t }^{ 2 }\\ dx=2tdt\\ 2\int _{ 0 }^{ 1 }{ \frac { { t }^{ 2 } }{ 1+t } dt } =2\int _{ 0 }^{ 1 }{ \left( t-1+\frac { 1 }{ 1+t }  \right) dt } =-1+2\ln { 2 } \\ =\ln { 2 } -\frac { 1 }{ 2 } \left( -1+2\ln { 2 }  \right) =\frac { 1 }{ 2 } \\ $$
A: An alternative way, by Feynman's trick:
$$ \mathfrak{I}=\int_{0}^{1}2x\log(1+x)\,dx=\left.\frac{d}{d\alpha}\int_{0}^{1}2x(1+x)^{\alpha}\,dx\right|_{\alpha=0^+} $$
and by writing $2x$ as $2(1+x)-2$ we get:
$$ \mathfrak{I}=2\left.\frac{d}{d\alpha}\left(\frac{2^{\alpha+2}-1}{\alpha+2}-\frac{2^{\alpha+1}-1}{\alpha+1}\right)\right|_{\alpha=0^+}$$
or
$$\mathfrak{I}=2\left[-\tfrac{3}{4}+2\log 2+1-2\log 2\right] = \frac{1}{2}.$$

Yet another way: $\log(1+x)=\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\,x^n$ leads to
$$ \mathfrak{I} = 2\sum_{n\geq 1}\frac{(-1)^{n+1}}{n(n+2)} = \sum_{n\geq 1}\left[\frac{(-1)^{n+1}}{n}-\frac{(-1)^{n+3}}{n+2}\right]$$
and this is a telescopic series.
