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

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lgbasalloteBest ResponseYou've already chosen the best response.0
this is very easy once you found the first step :P
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.1
hint please; i don't want to go on the track...what substitution?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
if i say the substitution then it's already solved :P that's the key haha
 one year ago

HeroBest ResponseYou've already chosen the best response.0
I didn't say anything relevant :P
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
i assume you found the substitution already..
 one year ago

HeroBest ResponseYou've already chosen the best response.0
I'm not saying anything :P
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.1
well, times the denominator and numerator by something? i dont want to go on the wrong track it will get messy
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
nope either :) just a simple u substitution
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
yup! yay :D now solve it >:))
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.1
lol, then its long division too lazy
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
hahaha =)) that's why it's an lgbariddle ;D
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
Take u =x^(1/6) and then go ahead with partial fraction . Tiring :P
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
Yeah. That's the only way to do it (I think).
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
ahh the beauty of lgbariddles >:))
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
No 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
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
Answer should come within 2 steps :D
 one year ago

wasiqssBest ResponseYou've already chosen the best response.0
shivam you have nnothing now :P
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
lol. I forgot du everywhere :P
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.3
"shivam you have nnothing now :P ">????
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
Is the first integral an arctan of something?
 one year ago

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

blockcolderBest ResponseYou've already chosen the best response.0
And then complete the square of the denominator and viola! A wild arctan appears!
 one year ago

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

shivam_bhallaBest ResponseYou've already chosen the best response.3
substitute u= x^(1/6) and voila :P
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
Oh, right. I thought the u^31 is in the denominator. Guess I should clean my glasses.
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
Yeah, that happens a lot with me. =))
 one year ago
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