evaluate the following integral. ∫14x ln(6x) dx

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Let $u = \ln(6x)$ and $dv=14x;dx$. Then $du=\frac{1}{x}dx$ and $v = 7x^{2}$ (since $\int 14x;dx=7x^{2}+C$).

Step2: Apply the formula

$\int 14x\ln(6x)dx=7x^{2}\ln(6x)-\int 7x^{2}\cdot\frac{1}{x}dx$.

Step3: Simplify the second integral

$\int 7x^{2}\cdot\frac{1}{x}dx=\int 7x;dx$. And $\int 7x;dx=\frac{7}{2}x^{2}+C$.

Answer:

$7x^{2}\ln(6x)-\frac{7}{2}x^{2}+C$