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 $$\frac13\tan^{-1}\left(\frac x3\right)-3 \log(1+x) + C$$
Best Answer
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 $A,B,C$