Answer
$$\frac{-1}{6 \left(4 x^{3}+3 x^{2}\right) }+c$$
Work Step by Step
Given $$\int \frac{2 x^{2}+x}{\left(4 x^{3}+3 x^{2}\right)^{2}} d x$$
Let
$$u=4 x^{3}+3 x^{2}\ \ \ \Rightarrow \ \ du=6(2x^2+x)dx $$
then
\begin{align*}
\int \frac{2 x^{2}+x}{\left(4 x^{3}+3 x^{2}\right)^{2}} d x&=\frac{1}{6}\int \frac{du}{u^{2}} \\
&=\frac{-1}{6u}+c\\
&=\frac{-1}{6 \left(4 x^{3}+3 x^{2}\right) }+c\\
\end{align*}
