$\int\left(\frac{x-3}{x^2+9}\right)^2 \mathrm{~d} x=$

Evaluate the integral $\int\left(\frac{x-3}{x^2+9}\right)^2 \mathrm{d}x$.

Expand the integrand and rewrite the numerator:

$\left(\frac{x-3}{x^2+9}\right)^2 = \frac{x^2 - 6x + 9}{(x^2+9)^2} = \frac{(x^2+9) - 6x}{(x^2+9)^2} = \frac{1}{x^2+9} - \frac{6x}{(x^2+9)^2}$.

The integral becomes:

$I = \int \frac{1}{x^2+9} \mathrm{d}x - \int \frac{6x}{(x^2+9)^2} \mathrm{d}x$.

The first integral is standard: $\int \frac{1}{x^2+9} \mathrm{d}x = \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right)$.

For the second, use substitution with $u = x^2+9$, $\mathrm{d}u = 2x \mathrm{d}x$:

$\int \frac{6x}{(x^2+9)^2} \mathrm{d}x = 3 \int u^{-2} \mathrm{d}u = -\frac{3}{u} = -\frac{3}{x^2+9}$.

Combining results:

$I = \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right) + \frac{3}{x^2+9} + C$.

This matches option C.