Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 55

$-\frac{1}{24} (3 x^{4}+9 x^{2})^{-4}+c.$

We have \begin{align*} \int \frac{\left(2 x^{3}+3 x\right) d x}{\left(3 x^{4}+9 x^{2}\right)^{5}}&= \frac{1}{6}\int \frac{\left(12 x^{3}+18 x\right) d x}{\left(3 x^{4}+9 x^{2}\right)^{5}}\\ &=\frac{1}{6} \left(12 x^{3}+18 x\right) (3 x^{4}+9 x^{2})^{-5}\\ &=-\frac{1}{24} (3 x^{4}+9 x^{2})^{-4}+c. \end{align*}