Answer
\[ = \frac{{ - \sqrt {1 - 4{x^3}} }}{3} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{2{x^2}}}{{\sqrt {1 - 4{x^3}} }}dx} \hfill \\
\hfill \\
set\,\,the\,\,substitution \hfill \\
\hfill \\
u = 1 - 4{x^3}\,\,\,\,then\,\,\,\,du = - 12{x^2}dx \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{{2{x^2}}}{{\sqrt {1 - 4{x^3}} }}\,dx} = - \frac{1}{6}\int_{}^{} { - 6 \times \frac{{2{x^2}}}{{\sqrt {1 - 4{x^3}} }}dx} \hfill \\
\hfill \\
= - \frac{1}{6}\int_{}^{} {\frac{{ - 12{x^2}}}{{\sqrt {1 - 4{x^3}} }}} \,dx \hfill \\
\hfill \\
substitute\,\,for\,\,x{\text{ and}}\,\,du \hfill \\
\hfill \\
- \frac{1}{6}\int_{}^{} {\frac{1}{{\sqrt u }}} = - \frac{1}{6}{\int_{}^{} u ^{ - \frac{1}{2}}}du \hfill \\
\hfill \\
integrate\,\, \hfill \\
\hfill \\
- \frac{1}{6}\,\left( {2{u^{\frac{1}{2}}}} \right) + C = \frac{{ - \sqrt u }}{3} + C \hfill \\
\hfill \\
replace\,\,u\,\,with\,\,\,u = 1 - 4{x^3} \hfill \\
\hfill \\
= \frac{{ - \sqrt {1 - 4{x^3}} }}{3} + C \hfill \\
\end{gathered} \]
