lgbasallote
  • lgbasallote
\(\large \mathbf{\color{maroon}{L} \color{violet}{G} \color{orange}{B} \color{darkblue}{A} \color{gold}{R} \color{brown}{I} \color{pink}{D} \color{purple}{D} \color{green}{L} \color{white}{E}}\) \(\LARGE \int \frac{1}{\sqrt x - \sqrt[3]{x}}dx\) Hint: Be Creative!
Mathematics
chestercat
  • chestercat
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lgbasallote
  • lgbasallote
lgbasallote
  • lgbasallote
this is very easy once you found the first step :P
Mimi_x3
  • Mimi_x3
hint please; i don't want to go on the track...what substitution?

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lgbasallote
  • lgbasallote
if i say the substitution then it's already solved :P that's the key haha
Hero
  • Hero
not hard
lgbasallote
  • lgbasallote
shhh hero :S
Hero
  • Hero
I didn't say anything relevant :P
lgbasallote
  • lgbasallote
i assume you found the substitution already..
Hero
  • Hero
I'm not saying anything :P
Mimi_x3
  • Mimi_x3
let u =x^(1/3) ?
lgbasallote
  • lgbasallote
haha no mimi ^_^
Mimi_x3
  • Mimi_x3
well, times the denominator and numerator by something? i dont want to go on the wrong track it will get messy
lgbasallote
  • lgbasallote
nope either :) just a simple u substitution
Mimi_x3
  • Mimi_x3
u= x^(1/2) ?
lgbasallote
  • lgbasallote
still no :p haha
Mimi_x3
  • Mimi_x3
u =x^(1/6) ?
lgbasallote
  • lgbasallote
yup! yay :D now solve it >:))
Mimi_x3
  • Mimi_x3
lol, then its long division too lazy
lgbasallote
  • lgbasallote
hahaha =)) that's why it's an lgbariddle ;D
Mimi_x3
  • Mimi_x3
Let, Hero do it.. :P
lgbasallote
  • lgbasallote
hmmm..
anonymous
  • anonymous
Take u =x^(1/6) and then go ahead with partial fraction . Tiring :P
blockcolder
  • blockcolder
Yeah. That's the only way to do it (I think).
lgbasallote
  • lgbasallote
ahh the beauty of lgbariddles >:))
anonymous
  • anonymous
No it is plum \[\int\limits_{}^{} \large \frac{6u^5} {u^2(u-1)}\] \[\large 6\int\limits_{}^{}\frac{u^3}{u-1} = \large 6\int\limits_{}^{}\frac{u^3-1^3}{u-1} +6\int\limits_{}^{}\frac{1}{u-1}\] Now continue :D
anonymous
  • anonymous
Answer should come within 2 steps :D
wasiqss
  • wasiqss
shivam you have nnothing now :P
anonymous
  • anonymous
lol. I forgot du everywhere :P
wasiqss
  • wasiqss
lol du= whatever :P
anonymous
  • anonymous
"shivam you have nnothing now :P "-->????
blockcolder
  • blockcolder
Is the first integral an arctan of something?
anonymous
  • anonymous
You got to be kidding me @blockcolder just apply u^3-1^3 = (u-1)(u^2+1+u) and cancel the numerator u-1 and denominator u-1
blockcolder
  • blockcolder
And then complete the square of the denominator and viola! A wild arctan appears!
anonymous
  • anonymous
@blockcolder \[\large 6\int\limits\limits_{}^{}{(u^2+1+u)}du +6\int\limits\limits_{}^{}\frac{1}{u-1}\] \[\large 6u^3/3 + 6u+ 3u^2 + 6\log(u-1)\]
anonymous
  • anonymous
substitute u= x^(1/6) and voila :P
blockcolder
  • blockcolder
Oh, right. I thought the u^3-1 is in the denominator. Guess I should clean my glasses.
anonymous
  • anonymous
LOL
blockcolder
  • blockcolder
Yeah, that happens a lot with me. =))

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