If `inte^(x)/(x^(n))dx` and `-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 given problem, we need to find \( k_2 - k_1 \) based on the provided integral equation. Let's break it down step by step.

### Step 1: Understand the integral
We start with the integral:
\[ I_n = \int \frac{e^x}{x^n} \, dx \]
According to the problem, this integral can be expressed as:
\[ I_n = -\frac{e^x}{k_1 x^{n-1}} + \frac{1}{k_2 - 1} I_{n-1} \]
where \( I_{n-1} = \int \frac{e^x}{x^{n-1}} \, dx \).

### Step 2: Apply Integration by Parts
We will use integration by parts to evaluate \( I_n \). Let's set:
- \( u = e^x \) (which we will differentiate)
- \( dv = x^{-n} \, dx \) (which we will integrate)

Using integration by parts:
\[ I_n = \int u \, dv = u v - \int v \, du \]
This gives us:
\[ I_n = e^x \int x^{-n} \, dx - \int \left( e^x \cdot \frac{d}{dx}(x^{-n}) \right) \, dx \]

### Step 3: Compute the integrals
The integral \( \int x^{-n} \, dx \) can be computed as:
\[ \int x^{-n} \, dx = \frac{x^{-n+1}}{-n+1} = \frac{x^{1-n}}{1-n} \]
Now, we differentiate \( x^{-n} \):
\[ \frac{d}{dx}(x^{-n}) = -n x^{-n-1} \]
Thus, we can write:
\[ I_n = e^x \cdot \frac{x^{1-n}}{1-n} - \int e^x \cdot (-n x^{-n-1}) \, dx \]
This simplifies to:
\[ I_n = \frac{e^x x^{1-n}}{1-n} + n I_{n-1} \]

### Step 4: Rearranging the expression
Now we can rearrange the expression for \( I_n \):
\[ I_n = \frac{e^x}{1-n} x^{1-n} + n I_{n-1} \]
This gives us:
\[ I_n = -\frac{e^x}{n-1} x^{n-1} + \frac{1}{n-1} I_{n-1} \]

### Step 5: Comparing coefficients
From the problem statement, we have:
\[ I_n = -\frac{e^x}{k_1 x^{n-1}} + \frac{1}{k_2 - 1} I_{n-1} \]
By comparing coefficients, we can identify:
\[ k_1 = n - 1 \]
\[ k_2 = n \]

### Step 6: 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:
\[ \boxed{1} \]