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Study23
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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
Study23 Group Title
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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Argonx16 Group TitleBest 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

Argonx16 Group TitleBest 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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