`int \ 1/x{loge^(e x)*loge^(e^2x) * loge^(e^3x)}dx`

Options

A. 0
B. 1
C. e
D. 1/e

Answer

To solve the integral \( \int \frac{1}{x} \log_e(e^x) \log_e(e^{2x}) \log_e(e^{3x}) \, dx \), we can follow these steps:

### Step 1: Simplify the logarithmic expressions
Using the property of logarithms, \( \log_e(a^b) = b \log_e(a) \), we can simplify the logarithmic terms in the integral:
\[ \log_e(e^x) = x, \quad \log_e(e^{2x}) = 2x, \quad \log_e(e^{3x}) = 3x \]
Thus, the integral becomes:
\[ \int \frac{1}{x} (x)(2x)(3x) \, dx \]

### Step 2: Combine the terms
Now, we can combine the terms inside the integral:
\[ \int \frac{1}{x} (6x^3) \, dx = \int 6x^2 \, dx \]

### Step 3: Integrate
Now, we can integrate \( 6x^2 \):
\[ \int 6x^2 \, dx = 6 \cdot \frac{x^3}{3} + C = 2x^3 + C \]

### Final Answer
Thus, the final result of the integral is:
\[ 2x^3 + C \]

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