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I am trying to find the antiderivative of $8t^{-{1/2}}$ but im just not understanding how to do this.

I saw someone get out of $t^{1/2}$ the answer $2t^{1/2}$ Can someone help me out?

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    Well, it is formally correct. And it is a useful strategy, I like to throw away constants and recover them at the end. It is also a useful approach to $xe^x$ say. Guess antiderivative is $xe^x$, check by differentiation, oops, but easy fix.2012-12-13

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Don't be worried about a negative exponent that's not $-1$: you apply the same rules as you would when integrating, say, $\displaystyle\;\int 8t^3 dt,\;$ for example.

Those rules, recall, are:

For any $n\neq -1$ and any number $b$, $\int b\cdot x^n dx \quad = \quad b\int x^n dx \quad =\quad b\cdot \frac{1}{(n + 1)} x^{(n+1)} + C \quad \text{(where $C$ is a constant)}.$
$\text{So, since}\;\;(-1/2) \neq -1, \quad \int 8t^{-1/2} dt = \frac{8}{\left(\frac12\right)} t^{\,(-\frac12)\, +\, 1)} + C \quad = \;\dots \;?$

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    Note that $\dfrac{8}{\frac12} = \dfrac22\cdot \dfrac{8}{\frac12} = 16$. And $t^{(-1/2) + 1} = t^{1/2} = \sqrt{t}$. That gives you $16t^{1/2} + C = 16\sqrt{t} + C$.2012-12-13
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Hint: $\int x^\alpha dx = \frac{1}{\alpha + 1} x^{\alpha + 1} + C$, as long as $\alpha \neq -1$. Also, note that $\int 8 t^{-\frac{1}{2}} dt = 8 \int t^{-\frac{1}{2}} dt$.