`int e^(x)""((x-1)/(x^(2)))dx` is equal to

Options

A. `(e^(x))/(x^(2)) +c`
B. `(-e^(x))/(x^(2))+c`
C. `(e^(x))/(x)+c`
D. `(-e^(x))/(x)+c`

Explanation

To solve the integral

\[ \int e^x \frac{x-1}{x^2} \, dx, \]

we can break it down into simpler parts.

### Step 1: Rewrite the Integral
We can rewrite the integrand as follows:

\[ \int e^x \left( \frac{x}{x^2} - \frac{1}{x^2} \right) \, dx = \int e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) \, dx. \]

### Step 2: Split the Integral
Now, we can split the integral into two separate integrals:

\[ \int e^x \frac{1}{x} \, dx - \int e^x \frac{1}{x^2} \, dx. \]

### Step 3: Use Integration by Parts
For the first integral, we will use integration by parts. Let:

- \( u = \frac{1}{x} \) which gives \( du = -\frac{1}{x^2} \, dx \),
- \( dv = e^x \, dx \) which gives \( v = e^x \).

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):

\[ \int e^x \frac{1}{x} \, dx = e^x \cdot \frac{1}{x} - \int e^x \left(-\frac{1}{x^2}\right) \, dx. \]

This simplifies to:

\[ \int e^x \frac{1}{x} \, dx = \frac{e^x}{x} + \int e^x \frac{1}{x^2} \, dx. \]

### Step 4: Combine the Integrals
Now, substituting back into our expression, we have:

\[ \int e^x \frac{x-1}{x^2} \, dx = \left( \frac{e^x}{x} + \int e^x \frac{1}{x^2} \, dx \right) - \int e^x \frac{1}{x^2} \, dx. \]

The integrals of \( \int e^x \frac{1}{x^2} \, dx \) cancel out, leaving us with:

\[ \int e^x \frac{x-1}{x^2} \, dx = \frac{e^x}{x} + C, \]

where \( C \) is the constant of integration.

### Final Result
Thus, the final result is:

\[ \int e^x \frac{x-1}{x^2} \, dx = \frac{e^x}{x} + C. \]