Evaluate the following integral using integration by parts. \[ 16 x \ln 3 x d x \]

To evaluate the integral \( \int 16 x \ln(3x) \, dx \) using integration by parts, we start by identifying our \( u \) and \( dv \). A good choice would be: Let \( u = \ln(3x) \) and \( dv = 16x \, dx \). Then we compute \( du \) and \( v \): 1. Differentiate \( u \): \[ du = \frac{1}{3x} \cdot 3 \, dx = \frac{1}{x} \, dx \] 2. Integrate \( dv \): \[ v = \int 16x \, dx = 8x^2 \] Now, by the integration by parts formula \( \int u \, dv = uv - \int v \, du \): Substituting our values, we have: \[ \int 16x \ln(3x) \, dx = 8x^2 \ln(3x) - \int 8x^2 \cdot \frac{1}{x} \, dx \] This simplifies to: \[ = 8x^2 \ln(3x) - \int 8x \, dx \] Now we integrate \( 8x \): \[ = 8x^2 \ln(3x) - 4x^2 + C \] Thus, the final answer is: \[ \int 16x \ln(3x) \, dx = 8x^2 \ln(3x) - 4x^2 + C \]