Answer
$\frac{1}{{225}}\left( {2\sqrt 3 - 3} \right)$
Work Step by Step
$$\eqalign{
& \int_0^{0.3} {\frac{x}{{{{\left( {9 - 25{x^2}} \right)}^{3/2}}}}} dx \cr
& {\text{This integral can be solved using the substitution method}} \cr
& {\text{Let }}u = 9 - 25{x^2} \cr
& du = - 50xdx,{\text{ }}dx = - \frac{1}{{50x}}du \cr
& {\text{The new limits of integration are:}} \cr
& x = 0.3 \to u = 9 - 25{\left( {0.3} \right)^2} = 6.75 \cr
& x = 0 \to u = 9 - 25{\left( 0 \right)^2} = 9 \cr
& {\text{Substituting}}{\text{, we obtain}} \cr
& \int_0^{0.3} {\frac{x}{{{{\left( {9 - 25{x^2}} \right)}^{3/2}}}}} dx = \int_9^{6.75} {\frac{x}{{{u^{3/2}}}}} \left( { - \frac{1}{{50x}}} \right)du \cr
& = - \frac{1}{{50}}\int_9^{6.75} {\frac{1}{{{u^{3/2}}}}} du \cr
& = - \frac{1}{{50}}\int_9^{6.75} {{u^{ - 3/2}}} du \cr
& = - \frac{1}{{50}}\left[ {\frac{{{u^{ - 1/2}}}}{{ - 1/2}}} \right]_9^{6.75} \cr
& = \frac{1}{{25}}\left[ {\frac{1}{{\sqrt u }}} \right]_9^{6.75} \cr
& = \frac{1}{{25}}\left( {\frac{1}{{\sqrt {6.75} }} - \frac{1}{{\sqrt 9 }}} \right) \cr
& = \frac{1}{{25}}\left( {\frac{{2\sqrt 3 - 3}}{9}} \right) \cr
& = \frac{1}{{225}}\left( {2\sqrt 3 - 3} \right) \cr} $$
