evaluate the following integral. ∫ 6x ln(3x) dx

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Let $u = \ln(3x)$ and $dv = 6x;dx$. Then $du=\frac{1}{x}dx$ and $v=\int6x;dx = 3x^{2}$.

Step2: Substitute into the formula

$\int6x\ln(3x)dx=3x^{2}\ln(3x)-\int3x^{2}\cdot\frac{1}{x}dx$.

Step3: Simplify the second integral

$\int3x^{2}\cdot\frac{1}{x}dx=\int3x;dx=\frac{3}{2}x^{2}+C$.

Step4: Write the final result

$\int6x\ln(3x)dx = 3x^{2}\ln(3x)-\frac{3}{2}x^{2}+C$.

Answer:

$3x^{2}\ln(3x)-\frac{3}{2}x^{2}+C$