Answer
$\displaystyle \frac{7x^{2}-9x+3}{(x^{2}+7)^{2}}=\frac{\mathrm{A}x+B}{x^{2}+7}+\frac{Cx+D}{(x^{2}+7)^{2}}$
Work Step by Step
The factor $(x^{2}+7)$ is a repeated quadratic factor,
see p.548. Repeated Prime Quadratic Factor In the Denominator
Include one partial fraction with a $linear$ numerator
for each $power$ of a prime, repeated quadratic factor in the denominator
Setup only:
$\displaystyle \frac{7x^{2}-9x+3}{(x^{2}+7)^{2}}=\frac{\mathrm{A}x+B}{x^{2}+7}+\frac{Cx+D}{(x^{2}+7)^{2}}$
