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The form of the partial fraction decomposition of a rational function is given below.

$\frac{10x^{2}-18x+36}{(x-3)(x^{2}+9)} = \frac{A}{x-3} + \frac{B x + C}{x^{2}+9}$
$A =$ $B =$ $C =$

Now evaluate the indefinite integral.

$\displaystyle\int \frac{10x^{2}-18x+36}{(x-3)(x^{2}+9)}\, dx =$ $+C$

You can earn partial credit on this problem.