## Thomas' Calculus 13th Edition

$\sqrt t+\dfrac{1}{t^3}+c$
Use formula $\int x^{a} dx=\dfrac{x^{(a+1)}}{(a+1)}+c$ where $c$ refers to a constant of proportionality. Then, we have $\int (\dfrac{1}{2} t^{-1/2}-3t^{-4}dt=\dfrac{1}{2}[\dfrac{t^{\dfrac{-1}{2}+1}}{\dfrac{-1}{2}+1}]+ (3)[\dfrac{t^{-4+1}}{-4+1}]+c$ Thus, $t^{(1/2)}+\dfrac{1}{t^3}+c=\sqrt t+\dfrac{1}{t^3}+c$