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 I know how to take an antiderivative. But this one's stumping me. I need it to finish a problem. What's the antiderivative of the square root of (8t + 3) ~Caitlyn

Hi Caitlyn,

Sometimes you can solve an antiderivative problem with an intelligent guess and then a modification to fix it if it doesnt work.

I expect that you know an antiderivative of $\sqrt x = x^{1/2}$ is $\frac23 x^{3/2},$ so maybe an antiderivative of $\sqrt{8t + 3} = (8t + 3)^{1/2}$ is $\frac23 (8t + 3)^{3/2}.$ You can check to see if it is correct by differentiating.

Let $f(t) = \frac23 (8t + 3)^{3/2}$ then using the power rule and the chain rule $f'(t) = \frac23 \times \frac32 (8t + 3)^{1/2} \times 8 = (8t + 3)^{1/2} \times 8.$ Hence $f(t) = \frac23 (8t + 3)^{3/2}$ gives an answer that is 8 times what you want so modify $f(t)$ by dividing it by 8, that is try $f(t) =\frac18 \times \frac23 (8t + 3)^{3/2} = \frac{2}{24} (8t + 3)^{3/2}.$ Differentiate this to see if you get $(8t + 3)^{1/2}.$

I hope this helps,
Penny

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