$\int \frac{d x}{\sqrt{x^2-x^5}}=\int \frac{d x}{x \sqrt{1-x^3}}=\int \frac{x^2 d x}{x^3 \sqrt{1-x^3}}$
$\begin{aligned} & 1-x^3=t \\ & \Rightarrow-3 x^2 d x=d t \\ & \Rightarrow-x^2 d x=\frac{-1}{3} d t \\ & \Rightarrow-\frac{-1}{3} \int \frac{d t}{(1-t) \sqrt{t}}\end{aligned}$
Let $\mathrm{t}=4^2$
$\mathrm{dt}=24 \mathrm{du}$
$=\frac{-2}{3} \int \frac{\mathrm{du}}{(1+4)(1-4)}=\left(\frac{-2}{3}\right)\left[\frac{1}{2} \int \frac{1}{1+4} \mathrm{du}+\frac{1}{2} \int \frac{1}{1-4} \mathrm{du}\right]$
$\begin{aligned} & =\frac{-1}{3}\left[\int \frac{1}{1+4} \mathrm{du}+\int \frac{1}{1-4} \mathrm{du}\right]+\mathrm{c} \\ & =\frac{-1}{3}\left[\log _{\mathrm{e}}|1+4|-\log _{\mathrm{e}}|1-4|\right]+\mathrm{c} \\ & =\frac{-1}{3}\left[\log _{\mathrm{e}}|1+\sqrt{\mathrm{t}}|-\log _{\mathrm{e}}|1-\sqrt{\mathrm{t}}|\right]+\mathrm{c} \\ & =\frac{-1}{3}\left[\log _{\mathrm{e}}\left|1+\sqrt{1-\mathrm{x}^3}\right|-\log _{\mathrm{e}}\left|1-\sqrt{1-\mathrm{x}^3}\right|\right]+\mathrm{c} \\ & =\frac{1}{3}\left[-\log _{\mathrm{e}}\left|1+\sqrt{1-\mathrm{x}^3}\right|+\log _{\mathrm{e}}\left|1-\sqrt{1-\mathrm{x}^3}\right|\right]+\mathrm{c}\end{aligned}$
$=\frac{1}{3}\left[\log _{\mathrm{e}}\left|\frac{1+\sqrt{1-\mathrm{x}^3}}{1+\sqrt{1-\mathrm{x}^3}}\right|\right]+\mathrm{c}$
$\Rightarrow f(x)=\frac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}} \Rightarrow f\left(\frac{1}{2}\right)=\frac{1-\sqrt{1-\frac{1}{8}}}{1+\sqrt{1-\frac{1}{8}}}$
$\Rightarrow \mathrm{f}\left(\frac{1}{2}\right)=\frac{1-\sqrt{\frac{7}{8}}}{1+\sqrt{\frac{7}{8}}}=\frac{\sqrt{8}-\sqrt{7}}{\sqrt{8}+\sqrt{7}}$
$\Rightarrow \mathrm{f}\left(\frac{1}{2}\right)=\frac{\sqrt{8}-\sqrt{7}}{\sqrt{8}+\sqrt{7}}$