The Second Fundamental Theorem of Calculus Question 4

For $x\gt 0$ , $\dfrac{d}{dx}\left(\int\limits_0^{3x}\ln(t^3+1) dt\right) =$

\ln(x^3+1)

\ln(27x^3+1)

3\ln(x^3+1)

3\ln(27x^3+1)

Approach

\frac{d}{dx}\left(\int\limits_0^{3x}\ln(t^3+1)  dt\right)

= \ln((3x)^3+1) \cdot \frac{d}{dx}(3x)

= \boxed{3\ln(27x^3+1)}