Calculus: Early Transcendentals, Chapter 4, 4.9, Section 4.9, Problem 15

The most general antiderivative G(t) of the function g(t) can be found using the following relation:
int g(t)dt = G(t) + c
int (1 + t + t^2)/(sqrt t)dt = int (1)/(sqrt t)dt + int (t)/(sqrt t)dt + int t^2/(sqrt t) dt
You need to use the following formula:
int t^(-n) dt = (t^(-n+1))/(-n+1)
int (1)/(sqrt t)dt= int t^(-1/2) dt = (t^(-1/2+1))/(-1/2+1) + c = 2t^(1/2) + c= 2sqrt t + c
int (t)/(sqrt t)dt= int (sqrt t*sqrt t)/(sqrt t)dt = int sqrt t dt = (t^(1/2+1))/(1/2+1) + c = (2/3)*t^(3/2) + c = (2/3)*tsqrt t + c
int t^2/(sqrt t) dt = int t^(2-1/2)dt = int t^(3/2)dt = (t^(3/2+1))/(3/2+1) + c
int t^2/(sqrt t) dt = (2/5)*t^(5/2) + c = (2/5)*t^2*sqrt t + c
Gathering all the results yields:
int (1 + t + t^2)/(sqrt t)dt = 2t^(1/2) + (2/3)*tsqrt t + (2/5)*t^2*sqrt t + c
Hence, evaluating the most general antiderivative of the function yields G(t) = 2t^(1/2) + (2/3)*tsqrt t + (2/5)*t^2*sqrt t + c.