Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 61

$\frac{1}{27} \tan(9t^3+1)+c.$

Since $(\tan(9t^3+1))'=27t^2\sec^2(9t^3+1)$, then we have \begin{align*} \int t^2 \sec^2(9t^3+1) d t &=\frac{1}{27}\int 27t^2 \sec^2(9t^3+1) d t \\ &=\frac{1}{27} \tan(9t^3+1)+c. \end{align*}