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anonymous
 4 years ago
\(\large \mathbf{\color{maroon}{L} \color{violet}{G} \color{orange}{B} \color{darkblue}{A} \color{gold}{R} \color{brown}{I} \color{pink}{D} \color{purple}{D} \color{green}{L} \color{white}{E}}\)
\(\LARGE \int \frac{1}{\sqrt x  \sqrt[3]{x}}dx\)
Hint: Be Creative!
anonymous
 4 years ago
\(\large \mathbf{\color{maroon}{L} \color{violet}{G} \color{orange}{B} \color{darkblue}{A} \color{gold}{R} \color{brown}{I} \color{pink}{D} \color{purple}{D} \color{green}{L} \color{white}{E}}\) \(\LARGE \int \frac{1}{\sqrt x  \sqrt[3]{x}}dx\) Hint: Be Creative!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is very easy once you found the first step :P

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.1hint please; i don't want to go on the track...what substitution?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if i say the substitution then it's already solved :P that's the key haha

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0I didn't say anything relevant :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i assume you found the substitution already..

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not saying anything :P

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.1well, times the denominator and numerator by something? i dont want to go on the wrong track it will get messy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nope either :) just a simple u substitution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yup! yay :D now solve it >:))

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.1lol, then its long division too lazy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hahaha =)) that's why it's an lgbariddle ;D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Take u =x^(1/6) and then go ahead with partial fraction . Tiring :P

blockcolder
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah. That's the only way to do it (I think).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahh the beauty of lgbariddles >:))

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No it is plum \[\int\limits_{}^{} \large \frac{6u^5} {u^2(u1)}\] \[\large 6\int\limits_{}^{}\frac{u^3}{u1} = \large 6\int\limits_{}^{}\frac{u^31^3}{u1} +6\int\limits_{}^{}\frac{1}{u1}\] Now continue :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Answer should come within 2 steps :D

wasiqss
 4 years ago
Best ResponseYou've already chosen the best response.0shivam you have nnothing now :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol. I forgot du everywhere :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"shivam you have nnothing now :P ">????

blockcolder
 4 years ago
Best ResponseYou've already chosen the best response.0Is the first integral an arctan of something?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You got to be kidding me @blockcolder just apply u^31^3 = (u1)(u^2+1+u) and cancel the numerator u1 and denominator u1

blockcolder
 4 years ago
Best ResponseYou've already chosen the best response.0And then complete the square of the denominator and viola! A wild arctan appears!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@blockcolder \[\large 6\int\limits\limits_{}^{}{(u^2+1+u)}du +6\int\limits\limits_{}^{}\frac{1}{u1}\] \[\large 6u^3/3 + 6u+ 3u^2 + 6\log(u1)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0substitute u= x^(1/6) and voila :P

blockcolder
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, right. I thought the u^31 is in the denominator. Guess I should clean my glasses.

blockcolder
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, that happens a lot with me. =))
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