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Mimi_x3

  • 2 years ago

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

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  1. Mimi_x3
    • 2 years ago
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    @wasiqss

  2. wasiqss
    • 2 years ago
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    lemme try

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

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

  5. wasiqss
    • 2 years ago
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    okay

  6. wasiqss
    • 2 years ago
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    ash dont solve its for me

  7. ash2326
    • 2 years ago
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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

  8. Mimi_x3
    • 2 years ago
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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

  9. wasiqss
    • 2 years ago
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    mishoo it was easy ~~~ i did it

  10. wasiqss
    • 2 years ago
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    just apply by parts

  11. Mimi_x3
    • 2 years ago
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    by parts? please elaborate.

  12. Mimi_x3
    • 2 years ago
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    i thought the first step is a substitution..

  13. wasiqss
    • 2 years ago
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    yeh take 1 as first functio

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

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

  16. wasiqss
    • 2 years ago
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    if it can be done easily why take substitution..

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

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

  19. Mimi_x3
    • 2 years ago
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    lol, but look at the answer on wolfram..

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

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

  22. wasiqss
    • 2 years ago
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    i can do better than wolfram :P

  23. Mimi_x3
    • 2 years ago
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    lol

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