question evaluate the integral. ∫ -6x^(-3) ln(5x) dx answer attempt 1 out of 2 submit answer

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Let $u = \ln(5x)$ and $dv=-6x^{-3}dx$. Then $du=\frac{1}{x}dx$ and $v = 3x^{-2}$.

Step2: Substitute into the formula

$\int - 6x^{-3}\ln(5x)dx=3x^{-2}\ln(5x)-\int3x^{-2}\cdot\frac{1}{x}dx$.

Step3: Simplify the second integral

$\int3x^{-2}\cdot\frac{1}{x}dx=\int3x^{-3}dx$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int3x^{-3}dx=3\cdot\frac{x^{-3 + 1}}{-3 + 1}=-\frac{3}{2}x^{-2}+C$.

Step4: Write the final result

$\int - 6x^{-3}\ln(5x)dx = 3x^{-2}\ln(5x)+\frac{3}{2}x^{-2}+C=\frac{3\ln(5x)}{x^{2}}+\frac{3}{2x^{2}}+C$.

Answer:

$\frac{3\ln(5x)}{x^{2}}+\frac{3}{2x^{2}}+C$