Answer
$2 t^{1/2} \ln t- 4 t^{1/2}+C$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
$\displaystyle \int t^{-1/2} \ln t \ dt=\ln t (2 t^{1/2})- \int (2 t^{1/2}) \times (\dfrac{1}{t}) \ dt$
or, $=2 t^{1/2} \ln t-2 \int ( t^{-1/2}) \ dt$
or, $=2 t^{1/2} \ln t- 4 t^{1/2}+C$
