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lgbasallote

  • 2 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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  1. lgbasallote
    • 2 years ago
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    @Ishaan94

  2. lgbasallote
    • 2 years ago
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    this is very easy once you found the first step :P

  3. Mimi_x3
    • 2 years ago
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    hint please; i don't want to go on the track...what substitution?

  4. lgbasallote
    • 2 years ago
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    if i say the substitution then it's already solved :P that's the key haha

  5. Hero
    • 2 years ago
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    not hard

  6. lgbasallote
    • 2 years ago
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    shhh hero :S

  7. Hero
    • 2 years ago
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    I didn't say anything relevant :P

  8. lgbasallote
    • 2 years ago
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    i assume you found the substitution already..

  9. Hero
    • 2 years ago
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    I'm not saying anything :P

  10. Mimi_x3
    • 2 years ago
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    let u =x^(1/3) ?

  11. lgbasallote
    • 2 years ago
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    haha no mimi ^_^

  12. Mimi_x3
    • 2 years ago
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    well, times the denominator and numerator by something? i dont want to go on the wrong track it will get messy

  13. lgbasallote
    • 2 years ago
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    nope either :) just a simple u substitution

  14. Mimi_x3
    • 2 years ago
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    u= x^(1/2) ?

  15. lgbasallote
    • 2 years ago
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    still no :p haha

  16. Mimi_x3
    • 2 years ago
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    u =x^(1/6) ?

  17. lgbasallote
    • 2 years ago
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    yup! yay :D now solve it >:))

  18. Mimi_x3
    • 2 years ago
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    lol, then its long division too lazy

  19. lgbasallote
    • 2 years ago
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    hahaha =)) that's why it's an lgbariddle ;D

  20. Mimi_x3
    • 2 years ago
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    Let, Hero do it.. :P

  21. lgbasallote
    • 2 years ago
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    hmmm..

  22. shivam_bhalla
    • 2 years ago
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    Take u =x^(1/6) and then go ahead with partial fraction . Tiring :P

  23. blockcolder
    • 2 years ago
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    Yeah. That's the only way to do it (I think).

  24. lgbasallote
    • 2 years ago
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    ahh the beauty of lgbariddles >:))

  25. shivam_bhalla
    • 2 years ago
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    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
    • 2 years ago
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    Answer should come within 2 steps :D

  27. wasiqss
    • 2 years ago
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    shivam you have nnothing now :P

  28. shivam_bhalla
    • 2 years ago
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    lol. I forgot du everywhere :P

  29. wasiqss
    • 2 years ago
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    lol du= whatever :P

  30. shivam_bhalla
    • 2 years ago
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    "shivam you have nnothing now :P "-->????

  31. blockcolder
    • 2 years ago
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    Is the first integral an arctan of something?

  32. shivam_bhalla
    • 2 years ago
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    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
    • 2 years ago
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    And then complete the square of the denominator and viola! A wild arctan appears!

  34. shivam_bhalla
    • 2 years ago
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    @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
    • 2 years ago
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    substitute u= x^(1/6) and voila :P

  36. blockcolder
    • 2 years ago
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    Oh, right. I thought the u^3-1 is in the denominator. Guess I should clean my glasses.

  37. shivam_bhalla
    • 2 years ago
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    LOL

  38. blockcolder
    • 2 years ago
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    Yeah, that happens a lot with me. =))

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