## lgbasallote 3 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!

1. lgbasallote

@Ishaan94

2. lgbasallote

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

3. Mimi_x3

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

4. lgbasallote

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

5. Hero

not hard

6. lgbasallote

shhh hero :S

7. Hero

I didn't say anything relevant :P

8. lgbasallote

i assume you found the substitution already..

9. Hero

I'm not saying anything :P

10. Mimi_x3

let u =x^(1/3) ?

11. lgbasallote

haha no mimi ^_^

12. Mimi_x3

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

13. lgbasallote

nope either :) just a simple u substitution

14. Mimi_x3

u= x^(1/2) ?

15. lgbasallote

still no :p haha

16. Mimi_x3

u =x^(1/6) ?

17. lgbasallote

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

18. Mimi_x3

lol, then its long division too lazy

19. lgbasallote

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

20. Mimi_x3

Let, Hero do it.. :P

21. lgbasallote

hmmm..

22. shivam_bhalla

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

23. blockcolder

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

24. lgbasallote

ahh the beauty of lgbariddles >:))

25. shivam_bhalla

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

26. shivam_bhalla

Answer should come within 2 steps :D

27. wasiqss

shivam you have nnothing now :P

28. shivam_bhalla

lol. I forgot du everywhere :P

29. wasiqss

lol du= whatever :P

30. shivam_bhalla

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

31. blockcolder

Is the first integral an arctan of something?

32. shivam_bhalla

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

33. blockcolder

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

34. shivam_bhalla

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

35. shivam_bhalla

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

36. blockcolder

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

37. shivam_bhalla

LOL

38. blockcolder

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