Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions I'm having issues with the Partial Fractions method:
$$\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )dx$$
I am doing this in a way I was taught, which somehow I feel is a bit different from other methods online.
So, first of all, we must factorize the denominator:
$$x^3+2x^2 = (x+2)\cdot x^2$$
Great. So now we write three fractions:
$$\frac{A}{x^2} + \frac{B}{x} + \frac{C}{x+2}$$
Eventually we conclude that
$$A(x+2)+B(x+2)(x)+C(x^2) = 5x^2+3x-2$$
So now we look at what happens when $x = -2$:
$$C = 12$$
When $x = 0$:
$$A = -1$$
And now we are missing $B$, but we can just pick an arbitrary number for $x$ like... $1$:
$$B = -1$$
We replace the values here:
$$\int \frac{-1}{x^2}dx + \int \frac{-1}{x}dx + \int \frac{12}{x+2}dx$$
Which results in
$$\frac{1}{x}-\ln(x)+12\ln(x+2)+K$$
But I fear the answer actually is
$$\frac{1}{x}+2\ln(x)+3\ln(x+2)+K$$
Can you tell me what did I do wrong, and what should I have done?

Once you arrive at $$A(x+2) + B(x+2)x + C(x^2) = 5x^2 + 3x - 2,$$ this equation must hold true for all $x$.  If you substitute $x = -2$, then  $$C(-2)^2 = 5(-2)^2 + 3(-2) - 2 = 5(4) - 6 - 2 = 12,$$ or $4C = 12$, or $$C = 3.$$  You seem to be under the impression that you can disregard any expressions involving $x$ on the left-hand side.  That is incorrect.
Next, substituting $x = 0$ gives $$A(0+2) = 5(0^2) + 3(0) - 2 = -2,$$ or $2A = -2$, or $$A = -1.$$  This you got right.  Now finally, if you were to choose $x = 1$, you would obtain $$A(1+2) + B(1+2)(1) + C(1^2) = 5(1^2) + 3(1) - 2 = 6,$$ and the left-hand side simplifies to $3A + 3B + C$.  To get $B$, we now have to substitute the values you previously obtained for $A$ and $C$:  $$3(-1) + 3B + 3 = 6,$$ or $$B = 2.$$  Therefore, the solution is $$\frac{5x^2 + 3x - 2}{x^2(x+2)} = \frac{-1}{x^2} + \frac{2}{x} + \frac{3}{x+2}.$$
