Question

\(\large \mathbf{\color{maroon}{L} \color{violet}{G} \color{orange}{B} \color{darkblue}{A} \color{gold}{R} \color{brown}{I} \color{pink}{D} \color{purple}{D} \color{green}{L} \color{white}{E}}\)

\(\LARGE \int \frac{1}{\sqrt x - \sqrt[3]{x}}dx\)

Hint: Be Creative!

Answer 1

@Ishaan94

Answer 2

this is very easy once you found the first step :P

Answer 3

hint please; i don't want to go on the track...what substitution?

Answer 4

if i say the substitution then it's already solved :P that's the key haha

Answer 5

not hard

Answer 6

shhh hero :S

Answer 7

I didn't say anything relevant :P

Answer 8

i assume you found the substitution already..

Answer 9

I'm not saying anything :P

Answer 10

let u =x^(1/3) ?

Answer 11

haha no mimi ^_^

Answer 12

well, times the denominator and numerator by something? i dont want to go on the wrong track it will get messy

Answer 13

nope either :) just a simple u substitution

Answer 14

u= x^(1/2) ?

Answer 15

still no :p haha

Answer 16

u =x^(1/6) ?

Answer 17

yup! yay :D now solve it >:))

Answer 18

lol, then its long division too lazy

Answer 19

hahaha =)) that's why it's an lgbariddle ;D

Answer 20

Let, Hero do it.. :P

Answer 21

hmmm..

Answer 22

Take u =x^(1/6)
and then go ahead with partial fraction . Tiring :P

Answer 23

Yeah. That's the only way to do it (I think).

Answer 24

ahh the beauty of lgbariddles >:))

Answer 25

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

Answer 26

Answer should come within 2 steps :D

Answer 27

shivam you have nnothing now :P

Answer 28

lol. I forgot du everywhere :P

Answer 29

lol du= whatever :P

Answer 30

"shivam you have nnothing now :P "-->????

Answer 31

Is the first integral an arctan of something?

Answer 32

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

Answer 33

And then complete the square of the denominator and viola! A wild arctan appears!

Answer 34

@blockcolder
\[\large 6\int\limits\limits_{}^{}{(u^2+1+u)}du +6\int\limits\limits_{}^{}\frac{1}{u-1}\]
\[\large 6u^3/3 + 6u+ 3u^2 + 6\log(u-1)\]

Answer 35

substitute u= x^(1/6) and voila :P

Answer 36

Oh, right. I thought the u^3-1 is in the denominator. Guess I should clean my glasses.

Answer 37

LOL

Answer 38

Yeah, that happens a lot with me. =))