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Find the most general antiderivative of the following function (note I have NOT learned about integrals yet). I'm confused about the chain rule and the antiderivative. The function: 2x+5(1x^2)^(1/2)
 one year ago
 one year ago
Find the most general antiderivative of the following function (note I have NOT learned about integrals yet). I'm confused about the chain rule and the antiderivative. The function: 2x+5(1x^2)^(1/2)
 one year ago
 one year ago

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Argonx16Best ResponseYou've already chosen the best response.1
First of all, for the sake of simplicity, lets pretend that antiderivatives are the same as integrals (they pretty much are). Also, when I say\[\int\limits_{}{} \frac {2x+5}{\sqrt{1x^2}}\] I mean "take the antiderivate of 2x+5(1x^2)^(1/2). So the first thing we notice is that we could rewrite the problem to \[\int\limits_ {}{} \frac {1}{\sqrt{1x^2}} * 2x+5\] If we look on the left hand side, we notice that we have the term \[\int\limits_ {}{} \frac {1}{1x^2}\] Which is the derivative of inverse sine \[\sin^{1}\]
 one year ago

Argonx16Best ResponseYou've already chosen the best response.1
EDIT. I totally mean to have a square root sign under the 1x^2 up there. My bad. So from that, we can conclude that inverse sine will have a part in this integral (antiderivative). So now that we know that inverse sine is part of the question, lets work through the integral again. \[\int\limits\limits\limits {}{} \frac{1}{\sqrt{1x^2}} (2x+5)\] The 1x^2 is the "driving function," or if you learn usubstitution later, the "U," meaning that if the "driving function/U" is responsible for the (2x+5) term per the chain rule. If this doesn't make any sense, just wait until you learn Usubstitution (or read about it yourself!)
 one year ago
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