Answer
$\ln (-t) (\dfrac{t^3}{3}-t) +\dfrac{t^3}{9}-t+C$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
$\displaystyle \int (t^2-t) \ln (-t) \ dt=\ln (-t) (\dfrac{t^3}{3}-t)- \int (\frac{t^3}{3}-t ) \times (-\dfrac{1}{t}) \ dt$
or, $=\ln (-t) (\dfrac{t^3}{3}-t) + \int (\dfrac{t^2}{3}-1 ) \ dt$
or, $=\ln (-t) (\dfrac{t^3}{3}-t) +\dfrac{t^3}{9}-t+C$
