The problem asks us to evaluate the indefinite integral $\int \frac{x+5}{x^2+x-2} \, dx$. The problem also provides a partial fraction decomposition, so we just need to evaluate the integral of the decomposed expression: $\int (\frac{2}{x-1} - \frac{1}{x+2}) \, dx$.

1. Problem Description

The problem asks us to evaluate the indefinite integral $\int \frac{x+5}{x^2+x-2} \, dx$. The problem also provides a partial fraction decomposition, so we just need to evaluate the integral of the decomposed expression: $\int (\frac{2}{x-1} - \frac{1}{x+2}) \, dx$.

2. Solution Steps

First, we observe that $x^2+x-2 = (x-1)(x+2)$. Thus, we can write

$$\frac{x+5}{x^2+x-2} = \frac{x+5}{(x-1)(x+2)}.$$

The given partial fraction decomposition is

$$\frac{x+5}{x^2+x-2} = \frac{2}{x-1} - \frac{1}{x+2}.$$

Now we integrate both sides with respect to $x$:

$$\int \frac{x+5}{x^2+x-2} \, dx = \int \left(\frac{2}{x-1} - \frac{1}{x+2}\right) \, dx.$$

We can split the integral:

$$\int \left(\frac{2}{x-1} - \frac{1}{x+2}\right) \, dx = \int \frac{2}{x-1} \, dx - \int \frac{1}{x+2} \, dx.$$

Now we can take out the constant 2 from the first integral:

$$2 \int \frac{1}{x-1} \, dx - \int \frac{1}{x+2} \, dx.$$

We know that $\int \frac{1}{u} \, du = \ln |u| + C$. Let $u = x-1$, then $du = dx$. Let $v = x+2$, then $dv = dx$. Thus,

$$$$

\int \frac{1}{x-1} \, dx = \int \frac{1}{u} \, du = \ln |u| + C_1 = \ln |x-1| + C_

1. $$

$$$$

\int \frac{1}{x+2} \, dx = \int \frac{1}{v} \, dv = \ln |v| + C_2 = \ln |x+2| + C_

2. $$

Substituting these results back into the original expression, we get

$$2 \ln |x-1| - \ln |x+2| + C,$$

where $C$ is the constant of integration. Using logarithm properties, we can write

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

3. Final Answer

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