## Thomas' Calculus 13th Edition

$-\dfrac{2}{t^2}-\dfrac{2}{3t^{3/2}}+C$
As we know $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$ Thus, we have$\int \dfrac{4 +\sqrt t}{t^3} dt=\int \dfrac{4}{t^{3}}+\dfrac{t^{1/2}}{t^3} dt$ or, $=4(\dfrac{1}{-2}t^{-2}+\dfrac{1}{-3/2}t^{-3/2}+C$ or, $=-\dfrac{2}{t^2}-\dfrac{2}{3t^{3/2}}+C$