Integrate $\int x \sqrt{2 - \sqrt{1-x^2}}dx $ it seems that integration by parts with some relation to substitution...

$$
\int x \sqrt{2-\sqrt{1-x^2}}
= \frac{2}{5} \sqrt{2-\sqrt{1-x^2}} \cdot \sqrt{1-x^2}+\frac{8}{15}\sqrt{2-\sqrt{1-x^2}}+c
$$

How can I get that?

Welcome to StackExchange! You may use LaTeX-Formulas by using dollar signs \$ x^2 \$ $\longrightarrow{}$ $x^2$.
As a first step I used the substitution $u:=\sqrt{2-x^2}$ which implies $ \frac{\mathrm{d}u}{\mathrm{d}x}=\frac{2x}{2\sqrt{1-x^2}}=\frac{x}{u}$. So we have to solve $\int \sqrt{2-u}\cdot u\ \mathrm{d}u$. This can be done by using partial integration (http://en.wikipedia.org/wiki/Integration_by_parts, use $v'=\sqrt{2-u}$), which yields $-2/15 (2-u)^ {3/2} (4+3 u)+c$. This should be equivalent to your result, if you replace $u$ by $\sqrt{1-x^2}$.
