Question

ol2: problem 1
(1 point)
use integration by parts to evaluate the integral.
\\( \int 4x \ln(2x) \\, dx = \square + c \\)
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Explanation:

Step1: Recall integration by parts formula

Integration by parts: $\int u \, dv = uv - \int v \, du$
Let $u = \ln(2x)$, $dv = 4x \, dx$

Step2: Compute $du$ and $v$

$du = \frac{1}{2x} \cdot 2 \, dx = \frac{1}{x} dx$
$v = \int 4x \, dx = 2x^2$

Step3: Apply integration by parts

$\int 4x\ln(2x) dx = 2x^2\ln(2x) - \int 2x^2 \cdot \frac{1}{x} dx$
Simplify the integral: $\int 2x^2 \cdot \frac{1}{x} dx = \int 2x dx$

Step4: Evaluate remaining integral

$\int 2x dx = x^2$

Step5: Combine terms

$\int 4x\ln(2x) dx = 2x^2\ln(2x) - x^2$

Answer:

$2x^2\ln(2x) - x^2$