anonymous
  • anonymous
need help with evaluating this integral.... (x^9)/(sqrt(3+x^5)) dx.....
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
I would recommend a u substitution of u = x^5+3
anonymous
  • anonymous
ok i started that....
anonymous
  • anonymous
How far did you get?

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anonymous
  • anonymous
then i got (1/5)du = x^4dx ....
anonymous
  • anonymous
Ok, though I prefer dx = du/(5x^4)
anonymous
  • anonymous
ok.... then
anonymous
  • anonymous
So then, what is x^5 in terms of u?
anonymous
  • anonymous
i dont know what you mean by that
anonymous
  • anonymous
u = x^5 + 3 so x^5 = ?
anonymous
  • anonymous
x^5 = u - 3 right?
anonymous
  • anonymous
u - 3
anonymous
  • anonymous
ok. Good. Now start substituting in your integral
anonymous
  • anonymous
so whats the x^9?... the denominator will be sqrt3?
anonymous
  • anonymous
The denominator will be \(\sqrt{u}\)
anonymous
  • anonymous
thats what i meant.///
anonymous
  • anonymous
the dx will be \[\frac{1}{5x^4}du\]
anonymous
  • anonymous
So what's left in the numerator( after you simplify)
anonymous
  • anonymous
idk.......
anonymous
  • anonymous
Cmon. \[\int \frac{x^9}{\sqrt{x^5+3}}dx = \int \frac{x^5 * x^4}{\sqrt{x^5+3}}dx \] \[\text{Let }u = x^5+3 \]\[\implies x^5 = u-3 \] \[\implies 5x^4dx = du \implies dx = \frac{1}{5x^4}du\] Therefore \[\int \frac{x^5 * x^4}{\sqrt{x^5+3}}dx = \int \frac{(u-3)*x^4}{\sqrt{u}} (\frac{1}{5x^4})du\]
anonymous
  • anonymous
And we can cancel the x^4 from top and bottom. Then split the fraction into two different fractions \[\frac{u}{\sqrt{u}} - \frac{3}{\sqrt{u}}\]
anonymous
  • anonymous
and integrate each one separately.
anonymous
  • anonymous
what happened to the 5 in the faraction
anonymous
  • anonymous
yes there should be a factor of 5 outside the integral , he most likely didnt mention it because any integral of a constant multiple is just the integral times that constant

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