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anonymous
 5 years ago
the indefinite integral i am given is (e^x  6x^2)/6... i need to evaluate it. please help!
anonymous
 5 years ago
the indefinite integral i am given is (e^x  6x^2)/6... i need to evaluate it. please help!

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nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Do you know the integral of e^x and the power rule?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i assume the integral is e^x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i need to use substitution, but i dont know what to substitute... i've tried it a lot already

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0You don't need to use any elevated integration techniques. If you want to prove that your result is right, just differentiate it and if you get the original integrand, you're fine.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what about the 6 in the denominator? how do i do that? is it just 6x?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0No, that is just a factor of 1/6 which you can pullout.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow.. i got it! thanks! i have another one tho if u are willing to help?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Pull that division into the difference and you should have two known integrals.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like (5/x)((2xe^x)/x)?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, and reduce that x in the second term.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok. then i would need substitution right?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0For what? You should know the integral of 1/x and e^x and that's all there is.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont know the 1/x one. is it natural log?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Exactly, sometimes it is defined this way \[\ln x := \int_1^x\frac 1 y \mathrm d y\] but if you have defined it as the inverse function of e^x you could use the inversedifferentiation rule.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got that one! thanks a lot. this is my last one... (x2)/(x^24x+4)^3..... that requires substitution correct?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, a little one :) but you should first observe that \[x^2  4x + 4 = (x2)^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i can cancel one of them right away?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would it turn into (x2)^6?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Sure, that is just one of the exponentiation rules.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then after the cancellation i am left with 1/(x2)^5?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now i substitute u=x2

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, but as du/dx = 1 if you are a little more experienced you could just do it in your head.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think it would be 1/((1/6)*(x2)^6)

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Not quite, maybe you should write it is \[(x2)^{5}\] instead to see it better.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this website free to use?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0It costs both of us our time, but you won't have to pay for it ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so y do ppl like u go around and answer questions?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0I love math and helping other people.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it seems strange that there is no charge for this service. or breech of security

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know what you mean by breech of security; but the idea of being is just that everyone can ask questions and give answers here. So if I had a question regarding my study, I could ask them too. Only that it seems there are a lot of highschool students here lately :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm... thats cool. i guess i'm a skeptic. but with quick responses like the ones u gave me i am becoming a believer
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