Answer:
[tex]6t + \frac{t^2}{2}+2/5t^{5/2} + C[/tex]
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
[tex]\int\limits {(6 + t+t^{3/2}}) \, dt\\[/tex]
Using the formula x^{n+1}/n+1
[tex]\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[/tex]
An antiderivative is a function that reverses what the derivative does.
Antiderivative is also known as integration.
Antiderivative of given function is [tex]6t+\frac{t^{2} }{2}+\frac{2}{5}t^{5/2}+C[/tex]
Antiderivative , [tex]\int\limits^ {}t^{n} \, dt=\frac{t^{n+1} }{n+1} +C[/tex]
So, [tex]\int\limits {(6+t+t^{2}/\sqrt{t} )} \, dt =\int\limits {(6+t+t^{3/2} )} \, dt\\\\=6t+t^{2}/2 + \frac{t^{3/2+1} }{3/2+1}[/tex]
[tex]\int\limits {g(t)} \, dt=6t+\frac{t^{2} }{2}+\frac{2}{5}t^{5/2}+C[/tex]
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