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Mimi_x3

\[\int\frac{1}{ln(x+1)} dx\]

  • one year ago
  • one year ago

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  1. Mimi_x3
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    @wasiqss

    • one year ago
  2. wasiqss
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    lemme try

    • one year ago
  3. Mimi_x3
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    Hint: similar to your other one.

    • one year ago
  4. Mimi_x3
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    But this is apparently an integral which does not have a close form..

    • one year ago
  5. wasiqss
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    okay

    • one year ago
  6. wasiqss
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    ash dont solve its for me

    • one year ago
  7. ash2326
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    I ain't solving, just for reference http://www.wolframalpha.com/input/?i=integral+1%2F%28ln+%28x%2B1%29%29+dx

    • one year ago
  8. Mimi_x3
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    this is like wasiqs previous integral which was apparently solveable http://www.wolframalpha.com/input/?i=integral+e^x%2F%28x%29

    • one year ago
  9. wasiqss
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    mishoo it was easy ~~~ i did it

    • one year ago
  10. wasiqss
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    just apply by parts

    • one year ago
  11. Mimi_x3
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    by parts? please elaborate.

    • one year ago
  12. Mimi_x3
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    i thought the first step is a substitution..

    • one year ago
  13. wasiqss
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    yeh take 1 as first functio

    • one year ago
  14. wasiqss
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    \[x/{(\ln(x+1)} - \int\limits_{}^{}x^2 +x\]

    • one year ago
  15. Mimi_x3
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    how about \(u= ln(x+1)\) => \(e^{u}= x+1\)

    • one year ago
  16. wasiqss
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    if it can be done easily why take substitution..

    • one year ago
  17. Mimi_x3
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    but this is an integral which does not have a closed form right?

    • one year ago
  18. wasiqss
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    i dont understand forms :P i just know how to solve the ugliest of integrals :P

    • one year ago
  19. Mimi_x3
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    lol, but look at the answer on wolfram..

    • one year ago
  20. wasiqss
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    i dont trust wolfram cause in two questions i pointed out its mistake and did what it cant do :P

    • one year ago
  21. Mimi_x3
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    lol are you sure that you're right though? :P

    • one year ago
  22. wasiqss
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    i can do better than wolfram :P

    • one year ago
  23. Mimi_x3
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    lol

    • one year ago
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