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

Options

A. `e^(x)((x-lnx)/(x))+c`
B. `e^(x)((x-lnx+1)/(x))+c`
C. `e^(x)((x-lnx)/(x^(2)))+c`
D. `e^(x)((x-lnx-1)/(x))+c`

Explanation

To solve the integral

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

we will follow these steps:

### Step 1: Simplify the integrand
We can rewrite the integrand by separating the terms:

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

This can be expressed as:

\[ e^x \left( \frac{1}{x} - \frac{1}{x^2} \right)(x - \ln x). \]

### Step 2: Rewrite the expression
Now, we can express the integral as:

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

### Step 3: Use substitution
Let us use the substitution:

\[ t = \frac{e^x}{x}. \]

Then, we need to find \( dt \):

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

Thus, we can express \( dx \) in terms of \( dt \):

\[ dx = \frac{x^2}{e^x (x - 1)} dt. \]

### Step 4: Substitute back into the integral
Now substituting back into the integral gives:

\[ \int \ln(t) \, dt, \]

where \( t = \frac{e^x}{x} \).

### Step 5: Integrate
The integral of \( \ln(t) \) is:

\[ t \ln(t) - t + C. \]

### Step 6: Substitute back for \( t \)
Now we substitute back for \( t \):

\[ \frac{e^x}{x} \ln\left(\frac{e^x}{x}\right) - \frac{e^x}{x} + C. \]

### Step 7: Simplify the expression
This simplifies to:

\[ \frac{e^x}{x} \left( x - \ln(x) - 1 \right) + C. \]

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

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