Answer
$-\dfrac{2}{t^2}-\dfrac{2}{3t^{3/2}}+C$
Work Step by Step
Given: $\int \dfrac{4 +\sqrt t}{t^3} dt$
Since, we have $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$
Thus, $\int \dfrac{4 +\sqrt t}{t^3} dt=\int \dfrac{4}{t^{3}}+\dfrac{t^{1/2}}{t^3} dt=4(\dfrac{1}{-2}t^{-2}+\dfrac{1}{-3/2}t^{-3/2}+C=-\dfrac{2}{t^2}-\dfrac{2}{3t^{3/2}}+C$
