Evaluate the integral. (Use C for the constant of integration. Remember to use absolute values where appropriate. \[ \int \frac{d x}{7 x \ln (x)-4 x} \]

To evaluate the integral \[ \int \frac{d x}{7 x \ln (x) - 4 x}, \] we will simplify the integrand first. Factoring out \( x \) from the denominator gives us: \[ 7 x \ln(x) - 4 x = x(7 \ln(x) - 4). \] Thus, we can rewrite the integral as: \[ \int \frac{dx}{x(7 \ln(x) - 4)}. \] Next, we can split this integral using the substitution method. Let \[ u = 7 \ln(x) - 4, \] which implies \[ du = \frac{7}{x} dx \quad \Rightarrow \quad dx = \frac{x}{7} du = \frac{e^{(u + 4)/7}}{7} du. \] Now we express \( x \) in terms of \( u \): \[ x = e^{(u + 4)/7}. \] Plugging this back into our integral, we have: \[ \int \frac{1}{x(7 \ln(x) - 4)} dx = \int \frac{1}{e^{(u + 4)/7}(u)} \cdot \frac{e^{(u + 4)/7}}{7} du = \int \frac{1}{7u} du. \] The integral of \( \frac{1}{u} \) is simply \( \ln |u| + C \): \[ \int \frac{1}{7u} du = \frac{1}{7} \ln |u| + C. \] Substituting back \( u = 7 \ln(x) - 4 \) gives: \[ \frac{1}{7} \ln |7 \ln(x) - 4| + C. \] Thus, the final answer is: \[ \int \frac{d x}{7 x \ln (x)-4 x} = \frac{1}{7} \ln |7 \ln(x) - 4| + C. \]