(i) $-\frac{2}{3} x(1-x)^{\frac{3}{2}}-\frac{4}{15}(1-x)^{\frac{5}{2}}+c_1$

(ii) $-\frac{2}{3}(1-x)^{\frac{3}{2}}+\frac{2}{5}(1-x)^{\frac{5}{2}}+c_2$

$(i)$

\begin{array}{l}  \int x(1-x)^{\frac{1}{2}} \mathrm{~d} x=x\left[-\frac{2}{3}(1-x)^{\frac{3}{2}}\right]-\int\left(-\frac{2}{3}\right)(1-x)^{\frac{3}{2}}(1) \mathrm{d} x\\
=-\frac{2}{3} x(1-x)^{\frac{3}{2}}+\frac{2}{3} \int(1-x)^{\frac{3}{2}} \mathrm{~d} x\\
=-\frac{2}{3} x(1-x)^{\frac{3}{2}}+\frac{2}{3}\left[\frac{(1-x)^{\frac{3}{2}+1}}{(-1)\left(\frac{3}{2}+1\right)}\right]+c_1\\
=-\frac{2}{3} x(1-x)^{\frac{3}{2}}-\frac{4}{15}(1-x)^{\frac{5}{2}}+c_1\\
\text { [Al] }
\end{array}

$(ii)$

\begin{aligned}
I &=\int x(1-x)^{\frac{1}{2}} \mathrm{~d} x \\
&=\int\left(1-u^2\right)\left(u^2\right)^{\frac{1}{2}}(-2 u) \mathrm{d} u \\
&=\int\left(1-u^2\right)(u)(-2 u) \mathrm{d} u \\
&=\int-2 u^2\left(1-u^2\right) \mathrm{d} u \\
&=\int\left(-2 u^2+2 u^4\right) \mathrm{d} u \\
&=-\frac{2 u^3}{3}+\frac{2 u^5}{5}+c_2 \\
&=-\frac{2}{3}(1-x)^{\frac{3}{2}}+\frac{2}{5}(1-x)^{\frac{5}{2}}+c_2
\end{aligned}

$(iii)$

\begin{array}{l}-\frac{2}{3} x(1-x)^{\frac{3}{2}}-\frac{4}{15}(1-x)^{\frac{5}{2}}+c_1-\left[-\frac{2}{3}(1-x)^{\frac{3}{2}}+\frac{2}{5}(1-x)^{\frac{5}{2}}+c_2\right] \\
=-\frac{2}{3} x(1-x)^{\frac{3}{2}}-\frac{4}{15}(1-x)^{\frac{5}{2}}+c_1+\frac{2}{3}(1-x)^{\frac{3}{2}}-\frac{2}{5}(1-x)^{\frac{5}{2}}-c_2 \\
=-\frac{2}{3} x(1-x)^{\frac{3}{2}}+\frac{2}{3}(1-x)^{\frac{3}{2}}-\frac{4}{15}(1-x)^{\frac{5}{2}}-\frac{2}{5}(1-x)^{\frac{5}{2}}+c_1-c_2 \\
=\frac{2}{3}(1-x)^{\frac{3}{2}}(-x+1)-(1-x)^{\frac{5}{2}}\left(\frac{2}{3}\right)+c_1-c_2 \\
=\frac{2}{3}(1-x)^{\frac{3}{2}}[(1-x)-(1-x)]+c_1-c_2 \\
=c_1-c_2 \text { (shown) [A1] }
\end{array}