Answer
\[ = \frac{\pi }{6}\]
Work Step by Step
\[\begin{gathered}
\int_{\frac{2}{{5\sqrt 3 }}}^{\frac{2}{5}} {\frac{{dx}}{{x\sqrt {25{x^2} - 1} }}} \hfill \\
\hfill \\
rewrite \hfill \\
\hfill \\
\int_{\frac{2}{{5\sqrt 3 }}}^{\frac{2}{5}} {\frac{{5dx}}{{5x\sqrt {{{\left( {5x\,} \right)}^2} - 1} }}} \hfill \\
\hfill \\
set\,\,u = 5x\,\,\,\,then\,\,\,du = 5dx \hfill \\
\hfill \\
x = \frac{2}{{5\sqrt 3 }}\,\,\,\,\,implies\,\,\,u = \frac{2}{{\sqrt 3 }} \hfill \\
\hfill \\
x = \frac{2}{5}\,\,\,\,\,\,\,\,\,\,\,\,\,implies\,\,u = 2 \hfill \\
\hfill \\
apply\,\,the\,\,\,substitution \hfill \\
\hfill \\
\int_{\frac{2}{{\sqrt 3 }}}^2 {\frac{{du}}{{u\sqrt {{u^2} - 1} }}} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \,\,\left[ {\operatorname{arcsec} \,u} \right]_{\frac{2}{{\sqrt 3 }}}^2 \hfill \\
\hfill \\
Fundamental\,\,theorem \hfill \\
\hfill \\
= \operatorname{arcsec} \,\left( 2 \right) - \operatorname{arcsec} \,\left( {\frac{2}{{\sqrt 3 }}} \right) \hfill \\
\hfill \\
= \frac{\pi }{3} - \frac{\pi }{6} \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
= \frac{\pi }{6} \hfill \\
\end{gathered} \]