Solve the integral: ∫ x^2+6x-3/(x+3)(x^2+2x+9)
I'm stuck. IDK if my setup is correct either.
∫A/(x+3) + ∫Bx+C/(x^2+2x+9)
F(x) = A ln(x+3) + ∫Bx+C/(x^2+2x+9)
X^2+6x-3=A(x^2+2x+9) + Bx+C(x+3)
X^2+6x-3=Ax^2+2Ax+9A+Bx^2+3Bx+Cx+3C
X^2+6x-3=x^2(A+B)+x(2A+3B+C)+3(3A+C)
I agree with your split-up of
(x^2+6x-3)/( (x+3)(x^2+2x+9) ) = A/(x-3) + (Bx+C)/(x^2 + 2x + 9)
notice the essential brackets.
so A(x^2+2x+9) + (Bx+C)(x-3) = x^2 + 6x - 3
usually I work from this line rather than expanding it
let x = -3 ---> A(9-6+9) + 0 = 9-18-3 ----> A = -1
let x = 0 ---> 9A + 3C = -3 -----> C = 2
let x = 1 ---> 12A + 4B + 4C = 4
3A + B + C = 1
-3 + B + 2 = 1 -----> B = 2
∫(x^2+6x-3)/(x+3)(x^2+2x+9)
= ∫ -1/(x-3) dx + ∫ (2x+2)/(x^2 + 2x + 9) dx
I am sure you can integrate that, notice that the derivative of x^2+2x+9 shows up very nicely as 2x+2, so you should be able to finish this in your head. Don't forget to add the constant.