anonymous
  • anonymous
integrate sqrt((x^3 -3)/x^11)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1327888347751:dw|
anonymous
  • anonymous
?
anonymous
  • anonymous
maybe we can take out that annoying \[x^{11}\] in the denominator and write is as \[\frac{1}{x^5}\sqrt{\frac{x^3-3}{x}}\]

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anonymous
  • anonymous
no that doesn't work. hmmm
anonymous
  • anonymous
the answer in the back of the book is 2/27 (1-3/x^3)^(3/2)+C
anonymous
  • anonymous
i got the answer from wolfram, but not the "show steps" and the answer makes it look like a u - sub so maybe we can arrange it to be one
anonymous
  • anonymous
it is supposed to be solved by u substitution
anonymous
  • anonymous
ok i have another idea. write it as \[\frac{1}{x^4}\sqrt{\frac{x^3-3}{x^3}}\] and then make \[u=\frac{x^3-3}{x^3}\]
anonymous
  • anonymous
that will do it.
anonymous
  • anonymous
okay, thankyou so much
anonymous
  • anonymous
let me try that
anonymous
  • anonymous
yw let me know if you run in to any problems, but a miracle will occur when you find \[du\] it will be just what you want
anonymous
  • anonymous
okay lol for sure.
anonymous
  • anonymous
it gives me (9/x^7) as du.. :/
anonymous
  • anonymous
check again, should be \[\frac{9}{x^4}\] as needed
anonymous
  • anonymous
ok i am gunna one more time.
anonymous
  • anonymous
\[\frac{d}{dx}\frac{x^3-3}{x^3}=\frac{x^3\times 3x^2-3x^2\times (x^3-3)}{x^6}\]
anonymous
  • anonymous
all the \[x^5\] terms go, you get \[\frac{9x^2}{x^6}=\frac{9}{x^4}\]
anonymous
  • anonymous
in fact you cannot get a power of 7 in the denominator, because \[(x^3)^2=x^6\]
anonymous
  • anonymous
oh shoot, i see my mistake, thankyou.. i was trying to do it with product rule
anonymous
  • anonymous
everyone thinks that is the easy way, but believe me it is not. even wolfram does it that way, but then you have to add the fractions an no one is good at that
anonymous
  • anonymous
yeah you are right, i will try not to use it, btw thankyou so much, ur a life saver
myininaya
  • myininaya
\[\frac{d}{dx}(\frac{x^3-3}{x^3})=\frac{d}{dx}(1-\frac{3}{x^3})\] quotient rule not needed but good work (the credit for this little part goes to JamesJ) Great work on figuring out the substitution, satellite. You're awesome! :)
anonymous
  • anonymous
\[\color{red}{\text{blush}}\]

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