Anti-derivative:
int(sqrt(1+sqrt(x))) dx
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anonymous · Answer

$$u = 1+\sqrt{x} \Rightarrow du = \frac{1}{2\sqrt{x}}dx\Rightarrow dx = 2(u-1)du$$\begin{eqnarray*}\int\sqrt{1+\sqrt{x}} dx &=& \int \sqrt{u}\cdot2(u-1)du\&=&2\int u^{3/2}du-2\int \sqrt{u}du\&=&2\frac{u^{5/2}}{5/2}-2\frac{u^{3/2}}{3/2}+K\&=&\frac{4}{5}u^{5/2}-\frac{4}{3}u^{3/2}+K\&=&4u^{3/2}\left(\frac{u}{2}-\frac{1}{3}\right)+K\&=&\frac{4}{15}u^{3/2}\left(3u-5\right)+K\&=&\frac{4}{15}\left(1+\sqrt{x}\right)^{3/2}(3(1+\sqrt{x})-5)+K\&=&\frac{4}{15}(1+\sqrt{x})^{3/2}(3\sqrt{x}-2)+K.\end{eqnarray*}