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.

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.
