Choose the correct answer of the given question
`int 1/(x(logx)^n) dx = ____+c`

Options

A. `(logx)^(1-n)/(1-n)`
B. `(logx)^(1+n)/(1+n)`
C. `(logx)^n/n`
D. `(logx)/(1-n)`

Answer

To solve the integral \( \int \frac{1}{x (\log x)^n} \, dx \), we will use the method of substitution. Here are the steps to arrive at the solution:

### Step 1: Choose a substitution
Let \( z = \log x \). Then, we differentiate \( z \) with respect to \( x \):
\[ dz = \frac{1}{x} \, dx \quad \Rightarrow \quad dx = x \, dz = e^z \, dz \]
because \( x = e^z \).

### Step 2: Rewrite the integral
Substituting \( z \) into the integral, we have:
\[ \int \frac{1}{x (\log x)^n} \, dx = \int \frac{1}{e^z z^n} \cdot e^z \, dz = \int \frac{1}{z^n} \, dz \]

### Step 3: Integrate
Now we need to integrate \( \frac{1}{z^n} \):
\[ \int z^{-n} \, dz = \frac{z^{-n+1}}{-n+1} + C = \frac{1}{1-n} z^{1-n} + C \quad \text{(for } n \neq 1\text{)} \]

### Step 4: Substitute back
Now, we substitute back \( z = \log x \):
\[ = \frac{1}{1-n} (\log x)^{1-n} + C \]

### Final Result
Thus, the final result of the integral is:
\[ \int \frac{1}{x (\log x)^n} \, dx = \frac{1}{1-n} (\log x)^{1-n} + C \]