What is `int(dx)/(x(1+lnx)^(n))` equal to ? `(n ne 1)`

Options

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

Answer

To solve the integral \( \int \frac{dx}{x(1+\ln x)^n} \) where \( n \neq 1 \), we can follow these steps:

### Step 1: Substitution
Let \( u = 1 + \ln x \). Then, we differentiate \( u \) to find \( dx \):
\[ \frac{du}{dx} = \frac{1}{x} \implies dx = x \, du \]
From the substitution \( u = 1 + \ln x \), we can express \( x \) in terms of \( u \):
\[ \ln x = u - 1 \implies x = e^{u-1} = \frac{e^u}{e} \]

### Step 2: Rewrite the Integral
Substituting \( u \) into the integral:
\[ \int \frac{dx}{x(1+\ln x)^n} = \int \frac{du}{(1+\ln x)^n} = \int \frac{du}{u^n} \]
Since \( dx = x \, du = \frac{e^u}{e} \, du \), we replace \( x \) in the integral:
\[ \int \frac{1}{\frac{e^u}{e} \cdot u^n} \cdot \frac{e^u}{e} \, du = \int \frac{e^u}{e \cdot u^n} \cdot \frac{e^u}{e} \, du = \int \frac{du}{u^n} \]

### Step 3: Integrate
Now we can integrate:
\[ \int u^{-n} \, du = \frac{u^{-n+1}}{-n+1} + C = \frac{1}{1-n} u^{1-n} + C \]

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

### Step 5: Final Answer
Thus, the final answer is:
\[ \int \frac{dx}{x(1+\ln x)^n} = -\frac{1}{n-1} (1 + \ln x)^{1-n} + C \]