Question

Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative.)
g(t) = 6 + t + t2/square root t

Answer 1

Answer:

$6t + \frac{t^2}{2}+2/5t^{5/2} + C$

Step-by-step explanation:

Given the expression;

g(t) = 6 + t + t²/√t

This can be rewritten as;

g(t) = 6 + t +t²/t^1/2

g(t) = 6 + t +t^{2-1/2}

g(t) = 6 + t +t^3/2

Integrate the result

$\int\limits {(6 + t+t^{3/2}}) \, dt$

Using the formula x^{n+1}/n+1

$\int\limits {(6 + t+t^{3/2}}) \, dt\\ = 6t + \frac{t^2}{2}+\frac{t^{3/2+1}}{3/2 + 1} \\ = 6t + \frac{t^2}{2}+\frac{t^{5/2}}{5/2} \\= 6t + \frac{t^2}{2}+2/5t^{5/2} + C$