Use integration by parts to evaluate the integral. \[ \int 4 x \ln (2 x) d x \]
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Beyond the Answer
To solve the integral \( \int 4 x \ln(2x) \, dx \) using integration by parts, we start by choosing our \( u \) and \( dv \) components. Let: - \( u = \ln(2x) \) (then \( du = \frac{1}{x} dx \)) - \( dv = 4x \, dx \) (then \( v = 2x^2 \)) Now applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \): 1. Calculate \( uv \): \[ uv = \ln(2x) \cdot 2x^2 = 2x^2 \ln(2x) \] 2. Calculate \( \int v \, du \): \[ du = \frac{1}{x} dx \rightarrow v \, du = 2x^2 \cdot \frac{1}{x} dx = 2x \, dx \] Therefore, \[ \int v \, du = \int 2x \, dx = x^2 \] Now we combine these results: \[ \int 4x \ln(2x) \, dx = 2x^2 \ln(2x) - \int x^2 \, dx \] Calculating the remaining integral: \[ \int x^2 \, dx = \frac{x^3}{3} \] Putting it all together: \[ \int 4x \ln(2x) \, dx = 2x^2 \ln(2x) - \frac{x^3}{3} + C \] So the final evaluation is: \[ \int 4x \ln(2x) \, dx = 2x^2 \ln(2x) - \frac{x^3}{3} + C \]
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