Enter (A,B,C) in order below if A, B, and C

$\frac{2x^2 + 3x - 5}{x(x^2 - 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}$

multiply both sides by $x(x^2-1)$:

$2x^2+3x-5=A(x-1)(x+1)+B(x)(x+1)+C(x)(x-1)$

Notice that if you set x to equal 0, 1, and -1, respectively, you can cancel out two of the 3 coefficients, which will make it easier to solve. First, set x = 0:

$-5=-A\\A=5$

Now set x = 1:

$2+3-5=2B\\B=0$

Lastly, set x = -1:

$2-3-5=2C\\ -6=2C\\C=-3$

Therefore, the answer is $\boxed{(5, 0, -3)}$