`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 \( I = \int \frac{e^x (x - 1)(x - \ln x)}{x^2} \, dx \), we can proceed step by step as follows:

### Step 1: Simplify the integrand
We can rewrite the integrand:
\[ I = \int \frac{e^x (x - 1)(x - \ln x)}{x^2} \, dx = \int e^x \left( \frac{x - 1}{x^2} \right) (x - \ln x) \, dx \]

### Step 2: Break down the expression
Next, we can express \( x - \ln x \) in a more manageable form:
\[ x - \ln x = \ln(e^x) - \ln x = \ln\left(\frac{e^x}{x}\right) \]
Thus, we can rewrite the integral as:
\[ I = \int e^x \left( \frac{x - 1}{x^2} \right) \ln\left(\frac{e^x}{x}\right) \, dx \]

### Step 3: Substitution
Let \( t = \frac{e^x}{x} \). Then, we differentiate \( t \):
\[ dt = \left(\frac{e^x}{x} \cdot (1 - \frac{1}{x})\right) \, dx = \frac{e^x (x - 1)}{x^2} \, dx \]
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 we can substitute \( t \) and \( dt \) back into the integral:
\[ I = \int \ln(t) \, dt \]

### Step 5: Integrate
The integral of \( \ln(t) \) is given by:
\[ \int \ln(t) \, dt = t \ln(t) - t + C \]

### Step 6: Substitute back for \( t \)
Now we substitute back \( t = \frac{e^x}{x} \):
\[ I = \frac{e^x}{x} \ln\left(\frac{e^x}{x}\right) - \frac{e^x}{x} + C \]

### Step 7: Simplify the expression
This can be simplified as:
\[ I = \frac{e^x}{x} \left( x - \ln(x) \right) - \frac{e^x}{x} + C \]
\[ I = \frac{e^x}{x} (x - 1 - \ln x) + C \]

### Final Result
Thus, the final result of the integral is:
\[ I = \frac{e^x}{x} (x - 1 - \ln x) + C \]