Answer
$\dfrac{3t^{4/3} \ln t}{4}-\dfrac{9t^{4/3}}{16}+C$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
$\displaystyle \int t^{1/3} \ln t \ dt=\ln t (\dfrac{3t^{4/3}}{4})- \int (\dfrac{3t^{4/3}}{4}) \times (\dfrac{1}{t}) \ dt$
or, $=\dfrac{3t^{4/3} \ln t}{4}- \dfrac{3}{4} (\dfrac{t^{4/3}}{4/3})+C$
or, $=\dfrac{3t^{4/3} \ln t}{4}-\dfrac{9t^{4/3}}{16}+C$
