## Calculus: Early Transcendentals (2nd Edition)

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