Answer
$F^{\prime}(x)=\displaystyle \frac{1}{x}$
Work Step by Step
Use the Second Fundamental Theorem of Calculus (Th 4.11)
$\displaystyle \frac{d}{dx}\left[\int_{a}^{x}f(t)dt\right] =f(x)$
Here, $F(x)$=$\displaystyle \int_{1}^{3x}f(t)dt,\qquad$ f(t) $=\displaystyle \frac{1}{t}, a=1$
With the substitution $u=3x, \displaystyle \frac{du}{dx}=3,$
we apply the chain rule
$F^{\prime}(x)=\displaystyle \frac{dF}{du}\cdot\frac{du}{dx}$
$=\displaystyle \frac{d}{du}[\int_{1}^{u}f(t)]\cdot 3$
$=f(u)\cdot 3$
$=\displaystyle \frac{1}{u}\cdot 3$
$=\displaystyle \frac{3}{3x}$
$=\displaystyle \frac{1}{x}$
