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 $$\frac13\tan^{-1}\left(\frac x3\right)-3 \log(1+x) + C$$
 A: Multiplying either sides by $(x+1)(x^2+9),$
$$x-3x^2-26=A(x^2+9)+(Bx+C)(x+1)$$
$$\implies-3x^2+x-26=x^2(A+B)+x(B+C)+9A+C$$
Compare the constants and the coefficients of $x,x^2$ of the above identity to form three simultaneous equations with three unknowns $A,B,C$  
A: 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 $A=-27/9$:
$$
         \frac{-26}{9} = A+\frac{C}{9} \implies C= 9\left(-\frac{26}{9}+\frac{27}{9}\right) = 1.
$$
A: An alternative technique called Heaviside cover-up makes the process simpler.
It gives $\begin{align}A = \frac{x-3x^2 - 26}{x^2 + 9}\Bigg|_{x=-1} =  -3\end{align}$
And then requires that $\dfrac{A}{x+1}$ be subtracted from both sides of the partial fraction decomposition identity. Hence, we get
$\begin{align} \frac{Bx+C}{x^2+9} &=\frac{x−3x^2−26}{(x+1)(x^2+9)} + \frac{3}{x+1}\\ &=\frac{x−3x^2−26 + 3x^2 + 27}{(x+1)(x^2+9)}\\&=\frac{1}{x^2+9}\end{align}$
$\therefore B = 0, C = 1$
The key feature of this technique is that it eliminates the need to solve a system of linear equations to obtain the coefficients, which can often be tedious.
See here and here for some more examples.
