(9t^2)(ln(t))dt at points 1 to e
So then, you calculated
$\displaystyle \int 9t^2 \ln t \cdot dt = 3t^3 \ln t - \int 3t^2 \cdot dt$
$\displaystyle \int 9t^2 \ln t \cdot dt = 3t^3 \ln t - t^3 + C$.
Hence
$\displaystyle \int _1 ^e 9t^2 \ln t \cdot dt = 3e^3 \ln e - e^3 - (3 \cdot 1^3 \ln 1 - 1^3)$,
by the Fundamental Theorem of Calculus. Just simplify that expression.