`int(1)/(x(1+logx)^(3))+c`

Options

A. `(1)/(2(1+logx)^(2))+c`
B. `(1)/(1+logx)+c`
C. `(-1)/(3(1+logx)^(3))+c`
D. `(-1)/(3(1+logx)^(3))+c`

Answer

To solve the integral \( \int \frac{1}{x(1 + \log x)^3} \, dx \), we will use the substitution method. Here’s a step-by-step solution:

### Step 1: Substitution
Let:
\[ y = 1 + \log x \]
Then, differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{x} \implies dy = \frac{1}{x} \, dx \]
This means:
\[ dx = x \, dy \]
Since \( x = e^{y-1} \) (from \( y = 1 + \log x \)), we can substitute \( x \) in terms of \( y \).

### Step 2: Rewrite the Integral
Substituting \( y \) and \( dx \) into the integral, we have:
\[ \int \frac{1}{x(1 + \log x)^3} \, dx = \int \frac{1}{x y^3} \cdot x \, dy = \int \frac{1}{y^3} \, dy \]

### Step 3: Integrate
Now we can integrate \( \frac{1}{y^3} \):
\[ \int \frac{1}{y^3} \, dy = \int y^{-3} \, dy = \frac{y^{-2}}{-2} + C = -\frac{1}{2y^2} + C \]

### Step 4: Substitute Back
Now, substitute back \( y = 1 + \log x \):
\[ -\frac{1}{2(1 + \log x)^2} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ -\frac{1}{2(1 + \log x)^2} + C \]