Calculus, 2ADV C4 2024 HSC 5 MC v1 What is \( {\displaystyle \int x(3 x^2+1)^4 d x} \) ? \( \dfrac{1}{30}(3x^2+1)^5+C \) \( \dfrac{1}{5}(3x^2+1)^5+C \) \( \dfrac{5}{6}(3x^2+1)^5+C \) \( \dfrac{6}{5}(3x^2+1)^5+C \) Show Answers Only \( A \) Show Worked Solution \[ \int x(3x^2+1)^4 dx\] \(=\dfrac{1}{5} \cdot \dfrac{1}{6x}x(3x^2+1)^5+C\) \(=\dfrac{1}{30}(3x^2+1)^5+C\) \( \Rightarrow A \) NOTE: Integrating by the reverse chain rule only works if some form of the derivative is already present outside of the brackets.