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