Question

evaluate the following integral.
int\frac{ln 13x}{x}dx
int\frac{ln 13x}{x}dx=square
(use parentheses to clearly denote the argument of each function.)

Explanation:

Step1: Use substitution

Let $u = \ln(13x)$. Then $du=\frac{1}{x}dx$.

Step2: Rewrite the integral

The integral $\int\frac{\ln(13x)}{x}dx$ becomes $\int udu$.

Step3: Integrate $u$

Using the power - rule for integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), for $n = 1$ in $\int udu$, we have $\frac{u^{2}}{2}+C$.

Step4: Substitute back $u$

Substituting $u=\ln(13x)$ back, we get $\frac{(\ln(13x))^{2}}{2}+C$.

Answer:

$\frac{(\ln(13x))^{2}}{2}+C$