Chain rule with a logarithmic function

Identify the outer function as $\ln(u)$ and the inner function as $u = x^2 + 3x$. The derivative of the outer function with respect to $u$ is $1/u$, and the derivative of the inner function with respect to $x$ is $2x+3$. By the chain rule, $f'(x) = \dfrac{1}{x^2+3x}\cdot(2x+3)$.

Now substitute $x=1$: the inner function value is $1^2+3(1) = 4$, and the inner derivative is $2(1)+3 = 5$. So $f'(1) = \dfrac{5}{4}$.