anonymous
  • anonymous
Hello. I'm stuck on an integral , which seems a bit difficult... I've got a link of wolframalpha where I followed the instructions .... but It seems something I don't understand... http://www.wolframalpha.com/input/?i=%28%28x%5E%280.5%29-1%29%5E3%29%2F%28x%29 I'll attach a photo too !
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Picture is in high resolution...
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anonymous
  • anonymous
Maybe du substitution has something to do with it.. but I need to know how ! :(
Zarkon
  • Zarkon
expand..simplify...integrate

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anonymous
  • anonymous
actually in wolframa there's a next subtitution we have to make... (before doing it..) but how does number 2 appear there? ... that's what I don't get !
.Sam.
  • .Sam.
@Zarkon I thought of expanding too, but its tedious, Leu u^2=x 2udu=dx \[\int\limits_{}^{}\frac{(u-1)^3}{u^2}(2u)du\] Simplify it
anonymous
  • anonymous
ok ...
anonymous
  • anonymous
thank you
anonymous
  • anonymous
\[\LARGE du=\frac{1}{2\sqrt{x}}dx \quad \quad ,\quad \sqrt x=u\] then... \[\LARGE du=\frac{1}{2u}dx \Longrightarrow 2u\cdot du=dx\] .. I got it ! Thank you !
anonymous
  • anonymous
dang that is a lot of extra to ing and fro ing for not much reward \[(a+b)^3=a^3+3a^2b+3ab^2+b^3\] you only have 4 terms \[(\sqrt{x}-1)^3=x^{\frac{3}{2}}-3x++3\sqrt{x}-1\] divide by x and it is an exercise in adding and subtracting exponents
anonymous
  • anonymous
i would find a substitution much more "tedious' but i guess it is a matter of taste

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