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!
Delete
Share
This Question is Closed
lgbasallote
Best Response
You've already chosen the best response.
0
@Ishaan94
lgbasallote
Best Response
You've already chosen the best response.
0
this is very easy once you found the first step :P
Mimi_x3
Best Response
You've already chosen the best response.
1
hint please; i don't want to go on the track...what substitution?
lgbasallote
Best Response
You've already chosen the best response.
0
if i say the substitution then it's already solved :P that's the key haha
Hero
Best Response
You've already chosen the best response.
0
not hard
lgbasallote
Best Response
You've already chosen the best response.
0
shhh hero :S
Hero
Best Response
You've already chosen the best response.
0
I didn't say anything relevant :P
lgbasallote
Best Response
You've already chosen the best response.
0
i assume you found the substitution already..
Hero
Best Response
You've already chosen the best response.
0
I'm not saying anything :P
Mimi_x3
Best Response
You've already chosen the best response.
1
let u =x^(1/3) ?
lgbasallote
Best Response
You've already chosen the best response.
0
haha no mimi ^_^
Mimi_x3
Best Response
You'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
lgbasallote
Best Response
You've already chosen the best response.
0
nope either :) just a simple u substitution
Mimi_x3
Best Response
You've already chosen the best response.
1
u= x^(1/2) ?
lgbasallote
Best Response
You've already chosen the best response.
0
still no :p haha
Mimi_x3
Best Response
You've already chosen the best response.
1
u =x^(1/6) ?
lgbasallote
Best Response
You've already chosen the best response.
0
yup! yay :D now solve it >:))
Mimi_x3
Best Response
You've already chosen the best response.
1
lol, then its long division too lazy
lgbasallote
Best Response
You've already chosen the best response.
0
hahaha =)) that's why it's an lgbariddle ;D
Mimi_x3
Best Response
You've already chosen the best response.
1
Let, Hero do it.. :P
lgbasallote
Best Response
You've already chosen the best response.
0
hmmm..
shivam_bhalla
Best Response
You've already chosen the best response.
3
Take u =x^(1/6)
and then go ahead with partial fraction . Tiring :P
blockcolder
Best Response
You've already chosen the best response.
0
Yeah. That's the only way to do it (I think).
lgbasallote
Best Response
You've already chosen the best response.
0
ahh the beauty of lgbariddles >:))
shivam_bhalla
Best Response
You've already chosen the best response.
3
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
shivam_bhalla
Best Response
You've already chosen the best response.
3
Answer should come within 2 steps :D
wasiqss
Best Response
You've already chosen the best response.
0
shivam you have nnothing now :P
shivam_bhalla
Best Response
You've already chosen the best response.
3
lol. I forgot du everywhere :P
wasiqss
Best Response
You've already chosen the best response.
0
lol du= whatever :P
shivam_bhalla
Best Response
You've already chosen the best response.
3
"shivam you have nnothing now :P "-->????
blockcolder
Best Response
You've already chosen the best response.
0
Is the first integral an arctan of something?
shivam_bhalla
Best Response
You've already chosen the best response.
3
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
Best Response
You've already chosen the best response.
0
And then complete the square of the denominator and viola! A wild arctan appears!
shivam_bhalla
Best Response
You've already chosen the best response.
3
@blockcolder
\[\large 6\int\limits\limits_{}^{}{(u^2+1+u)}du +6\int\limits\limits_{}^{}\frac{1}{u-1}\]
\[\large 6u^3/3 + 6u+ 3u^2 + 6\log(u-1)\]
shivam_bhalla
Best Response
You've already chosen the best response.
3
substitute u= x^(1/6) and voila :P
blockcolder
Best Response
You've already chosen the best response.
0
Oh, right. I thought the u^3-1 is in the denominator. Guess I should clean my glasses.
shivam_bhalla
Best Response
You've already chosen the best response.
3
LOL
blockcolder
Best Response
You've already chosen the best response.
0
Yeah, that happens a lot with me. =))