If `I_n= inte^(x)/(x^(n))dx= (-e^(x))/(k_(1)x^(n-1))+1/(k_(2)-1)I_(n-1)`, then `(k_(2)-k_(1))` is equal to:

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to evaluate the integral \( I_n = \int \frac{e^x}{x^n} \, dx \) and compare it with the given equation:

\[ I_n = -\frac{e^x}{k_1 x^{n-1}} + \frac{1}{k_2 - 1} I_{n-1} \]

### Step 1: Apply Integration by Parts
We will use integration by parts where we let:
- \( u = e^x \) (which implies \( du = e^x \, dx \))
- \( dv = \frac{1}{x^n} \, dx \) (which implies \( v = -\frac{1}{(n-1)x^{n-1}} \))

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

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

This simplifies to:

\[ I_n = -\frac{e^x}{(n-1)x^{n-1}} + \frac{1}{(n-1)} \int \frac{e^x}{x^{n-1}} \, dx \]

### Step 2: Recognize the Integral
Notice that the remaining integral is \( I_{n-1} \):

\[ I_n = -\frac{e^x}{(n-1)x^{n-1}} + \frac{1}{(n-1)} I_{n-1} \]

### Step 3: Compare with the Given Equation
Now, we compare this result with the given equation:

\[ I_n = -\frac{e^x}{k_1 x^{n-1}} + \frac{1}{k_2 - 1} I_{n-1} \]

From our expression, we can identify:
- \( k_1 = n - 1 \)
- \( \frac{1}{k_2 - 1} = \frac{1}{(n-1)} \)

### Step 4: Solve for \( k_2 \)
From \( \frac{1}{k_2 - 1} = \frac{1}{(n-1)} \), we can solve for \( k_2 \):

\[ k_2 - 1 = n - 1 \implies k_2 = n \]

### Step 5: Calculate \( k_2 - k_1 \)
Now we can find \( k_2 - k_1 \):

\[ k_2 - k_1 = n - (n - 1) = 1 \]

### Final Answer
Thus, the value of \( k_2 - k_1 \) is:

\[ \boxed{1} \]