Answer
$-\frac{1}{9 }(x^3+3x)^{-3}+c.$
Work Step by Step
Since $u=x^3+3x$, then $du=(3x^2+3)dx=3(x^2+1)dx$, and hence, we have
$$
\int \frac{x^2+1}{(x^3+3x)^4}dx= \frac{1}{3}\int \frac{du}{u^4}=-\frac{1}{9 }u^{-3}+c\\
=-\frac{1}{9 }(x^3+3x)^{-3}+c.
$$
