What is the value of the integral `int_1^pi e^x (1/x-1/x^2)dx`

Options

A. `e(e/2-1)`
B. `e(e-1)`
C. `e-1/e`
D. 0

Answer

To solve the integral \( I = \int_1^{\pi} e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) dx \), we can simplify the expression inside the integral and apply integration techniques.

### Step 1: Rewrite the Integral
We can rewrite the integral as:
\[ I = \int_1^{\pi} e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) dx = \int_1^{\pi} e^x \cdot \frac{1}{x} dx - \int_1^{\pi} e^x \cdot \frac{1}{x^2} dx \]

### Step 2: Identify the First Integral
Let’s denote:
\[ I_1 = \int_1^{\pi} e^x \cdot \frac{1}{x} dx \]
and
\[ I_2 = \int_1^{\pi} e^x \cdot \frac{1}{x^2} dx \]
Thus, we have:
\[ I = I_1 - I_2 \]

### Step 3: Solve \( I_1 \)
For \( I_1 \), we can use integration by parts. Let:
- \( u = \frac{1}{x} \) so that \( du = -\frac{1}{x^2} dx \)
- \( dv = e^x dx \) so that \( v = e^x \)

Using integration by parts:
\[ I_1 = \left[ \frac{e^x}{x} \right]_1^{\pi} - \int_1^{\pi} e^x \left(-\frac{1}{x^2}\right) dx \]
This simplifies to:
\[ I_1 = \left[ \frac{e^{\pi}}{\pi} - \frac{e^1}{1} \right] + I_2 \]

### Step 4: Substitute \( I_1 \) in \( I \)
Now substituting \( I_1 \) back into our expression for \( I \):
\[ I = \left[ \frac{e^{\pi}}{\pi} - e \right] + I_2 - I_2 \]
This simplifies to:
\[ I = \frac{e^{\pi}}{\pi} - e \]

### Step 5: Final Result
Thus, the value of the integral \( I \) is:
\[ I = \frac{e^{\pi}}{\pi} - e \]