anonymous
  • anonymous
indefinite integral of (2/cubed root of x) - 3 cubed root of x^2
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
3/2*x^(4/3)-9/5*x^(5/3)+C using the power rule for integrals.
anonymous
  • anonymous
I need some clarification on converting radical in denominator to numerator in exponential form
anonymous
  • anonymous
I always get confused. so if you could list a few of them and show me I will be able to apply the power rule for intgrals thanks

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anonymous
  • anonymous
Ok I think I might have got it...except the 9/5 how did you get that?
anonymous
  • anonymous
shouldn't it be 3/5 * x ^ (5/3)?
anonymous
  • anonymous
Hold on. We're trying to integrate 2*x^(-1/3)-3*x^(2/3). So... using integral rule, 2*1/(1-1/3)*x^(2/3)-3*1/(1+2/3)*x^(5/3)= 3x^(2/3)-9/5*x^(5/3)
anonymous
  • anonymous
+C. I did make a mistake initially.
anonymous
  • anonymous
\[2/ \sqrt[3]{x} - \sqrt[3]{x^2}\]
anonymous
  • anonymous
this is my question
anonymous
  • anonymous
Yes. I know that. Haha.
anonymous
  • anonymous
3x^(2/3)-3/5*x^(5/3)+C. Your 3 threw me off.
anonymous
  • anonymous
sorry about that.. now it totally makes sense, thanks
anonymous
  • anonymous
Fan me then. :D
anonymous
  • anonymous
can I do one and you check and see if still understand
anonymous
  • anonymous
\[dt/\sqrt[3]{t}\]
anonymous
  • anonymous
so i get 3/2* t^2/3 +c
anonymous
  • anonymous
Yup. :)
anonymous
  • anonymous
i also get confused about ln|x| stuff
anonymous
  • anonymous
example (2x^4 - x)/ x^3 dx = I get x^2 +x^-1 +c
anonymous
  • anonymous
Ask a new question about this preferably. I don't wanna clutter one question up.
anonymous
  • anonymous
ok I just want to know why can't I write ln|x| for x^-1
anonymous
  • anonymous
You should be able to?
anonymous
  • anonymous
The integral of x^-1 is ln|x|+C
anonymous
  • anonymous
ok.. that's all I want to know.. it's sometimes text book shows one way and the other time it shows ln |x| without any explanation
anonymous
  • anonymous
I am good on this one now thanks again
anonymous
  • anonymous
It should be ln|x| because you can't take a log of a negative number.

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