Partial fraction decomposition of a rational function The form of the partial fraction decomposition of a rational function is given below. 
$$\frac{x−3x^2−26}{(x+1)(x^2+9)} = \frac{A}{x+1}+ \frac{Bx+C}{x^2+9}$$
What are the values of $A,B$ and $C$?
So can someone explain how to get the answer. I'm not sure how I got to the answer but I know $B=0$ and $C=1$
I also found the indefinite integral, which equals $$\frac13\tan^{-1}\left(\frac x3\right)-3 \log(1+x) + C$$

Multiply by $x+1$ and let $x\rightarrow -1$ to obtain
$$
           \frac{-1-3-26}{(-1)^{2}+9}= A \implies A=-3.
$$
Multiply both sides by $x$, let $x\rightarrow\infty$, and use $A=-3$ to obtain
$$
            -3 = A + B \implies B=0.
$$
Finally, evaluate at $x=0$, and use $A=-27/9$:
$$
         \frac{-26}{9} = A+\frac{C}{9} \implies C= 9\left(-\frac{26}{9}+\frac{27}{9}\right) = 1.
$$
