Yes. Your answer is indeed correct. However, you need not have performed the integral. You could resort to Leibniz integral rule (Fundamental theorem of Calculus in disguise) for this. The rule goes as follows. If $F(x) = \int_{a(x)}^{b(x)} f(x,t) dt,$ then $F'(x) = \int_{a(x)}^{b(x)} \dfrac{\partial f(x,t)}{\partial x} dt + f(x,b(x)) \dfrac{db(x)}{dx} - f(x,a(x)) \dfrac{da(x)}{dx}$ In your case, $f(x,t) = \dfrac1t$, $a(x)=1$ and $b(x) = 3x$. Hence, we find that the first and last terms in the above equality are zero. The only non-zero term is the second term i.e. $F'(x) = f(x,b(x)) \dfrac{db(x)}{dx} = \dfrac1{3x} \times 3 = \dfrac1x$ Leibniz integral rule is especially handy when integrating $f(x,t)$ i.e. computing $\displaystyle \int_{a(x)}^{b(x)} f(x,t) dt$ explicitly is difficult.