The integral

$I=\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^{4}+3 x^{2}+1\right)^{4}} d x=\int \frac{3 x^{13}+2 x^{11}}{x^{16}\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{4}} d x$

$I=\int \frac{\frac{3}{x^{3}}+\frac{2}{x^{5}}}{\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{4}} d x$

Let $\quad 2+\frac{3}{x^{2}}+\frac{1}{x^{4}}=\mathrm{t},-2\left(\frac{3}{x^{3}}+\frac{2}{x^{5}}\right) d x=d t$

Then, $\quad I=\int \frac{-\frac{d t}{2}}{t^{4}}=-\frac{1}{2} \frac{t^{-4+1}}{-4+1}+C$

$I=\frac{-1}{2} \times \frac{1}{(-3)} \frac{1}{\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{3}}+C$

$I=\frac{1}{6} \frac{x^{12}}{\left(2 x^{4}+3 x^{2}+1\right)^{3}}+C$