Integrate (2x2+5x-1)/(x3+x2-2x) dx

The denominator is $x^3 + x^2 - 2x$.

$$x^3 + x^2 - 2x = x(x^2 + x - 2)$$

Factor the quadratic term $x^2 + x - 2$:

$$x^2 + x - 2 = (x+2)(x-1)$$

So, the fully factored denominator is:

$$x(x+2)(x-1)$$

We express the integrand as a sum of simpler fractions:

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

Multiply both sides by $x(x+2)(x-1)$:

$$2x^2 + 5x - 1 = A(x+2)(x-1) + Bx(x-1) + Cx(x+2)$$

To find A, set $x=0$:

$$2(0)^2 + 5(0) - 1 = A(0+2)(0-1)$$ $$-1 = A(2)(-1)$$ $$-1 = -2A \implies A = \frac{1}{2}$$

To find B, set $x=-2$:

$$2(-2)^2 + 5(-2) - 1 = B(-2)(-2-1)$$ $$2(4) - 10 - 1 = B(-2)(-3)$$ $$8 - 10 - 1 = 6B$$ $$-3 = 6B \implies B = -\frac{3}{6} = -\frac{1}{2}$$

To find C, set $x=1$:

$$2(1)^2 + 5(1) - 1 = C(1)(1+2)$$ $$2 + 5 - 1 = C(1)(3)$$ $$6 = 3C \implies C = 2$$

$$= \frac{1}{2} \int \frac{1}{x} \, dx - \frac{1}{2} \int \frac{1}{x+2} \, dx + 2 \int \frac{1}{x-1} \, dx$$ $$= \frac{1}{2} \ln|x| - \frac{1}{2} \ln|x+2| + 2 \ln|x-1| + C$$

$$= \frac{1}{2} \left( \ln|x| - \ln|x+2| \right) + \ln|(x-1)^2| + C$$ $$= \frac{1}{2} \ln \left| \frac{x}{x+2} \right| + \ln|(x-1)^2| + C$$

The final answer is $\boxed{\frac{1}{2} \ln \left| \frac{x}{x+2} \right| + \ln|(x-1)^2| + C}$.