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monroe17 Group Title

integral of ((lnx)^5)/(5x)dx please help step by step.. it's a practice problem for my exam this coming week.. and I need help reviewing because I've forgotten basically how to do these problems

  • one year ago
  • one year ago

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  1. Spacelimbus Group Title
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    This is a problem you want to solve by using the method of substitution. Got any idea how to do that?

    • one year ago
  2. monroe17 Group Title
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    u substitution..? I don't know how to do it.. it confuses me. the only thing i know is u= lnx and du= 1/x ?

    • one year ago
  3. Spacelimbus Group Title
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    First I recommend you to rewrite the integral, like this: \[\large \frac{1}{5}\int (\ln x)^5 \cdot \frac{1}{x}dx \]

    • one year ago
  4. Spacelimbus Group Title
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    Yes, perfect @monroe17 !! You're pretty close from there.

    • one year ago
  5. monroe17 Group Title
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    wait why and how did you rewrite it like that?

    • one year ago
  6. Spacelimbus Group Title
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    Well, first I did factor out the constants, then I just rewrote it to make the substitution more obvious. The rules are just algebraic. (Providing I understood your problem correctly) \[ \large \frac{(\ln x)^5}{5x}= (\ln x)^5 \cdot \frac{1}{5x} \]

    • one year ago
  7. monroe17 Group Title
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    question: (this may be a stupid question too) how'd you get 1/5 factored out?

    • one year ago
  8. Spacelimbus Group Title
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    It's not too obvious if you're new to integrals, but it's a general rule of integration, you're integrating in the dimension / direction of \(dx\) therefore all numbers \( \in \mathbb{R} \) can be factored out. This will make your integral look 'easier' Or at least you wont be confused by the numbers anymore. Of course you must multiply them back in at the end. However, to some people this seems like uneasy extra work and they just skip that step, which is perfectly fine too.

    • one year ago
  9. monroe17 Group Title
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    did it come from the 1/5x part?

    • one year ago
  10. Spacelimbus Group Title
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    exactly, I just thought the integral would look nicer without the unnecessary 5 in the denominator.

    • one year ago
  11. monroe17 Group Title
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    but then why did it stay 1/x ... why wouldn't it be x?

    • one year ago
  12. Spacelimbus Group Title
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    as an example:\[\Large \int \frac{1}{5x}dx = \frac{1}{5} \int \frac{1}{x}dx \]

    • one year ago
  13. Spacelimbus Group Title
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    Just a general law of multiplication \[ \large \frac{a}{b} \cdot \frac{c}{d}= \frac{ac}{bd} \]

    • one year ago
  14. monroe17 Group Title
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    oh lol.. okay

    • one year ago
  15. Spacelimbus Group Title
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    Most teachers provide that step, to some students - new to calculus - if you show them the integrals I posted above, they would say that they don't know how to integrate the first one, but the second one. It's more like to put empathy to the general structure of it, most of the time that's all what mathematics is about (-:

    • one year ago
  16. monroe17 Group Title
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    oh okayy.. so now do i find the anti derivative?

    • one year ago
  17. Spacelimbus Group Title
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    Use the substitution you have already figured out yourself, if you plug them into correctly you will see that it dramatically simplifies this devilish looking integral.

    • one year ago
  18. monroe17 Group Title
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    lol! okay :) give me a second

    • one year ago
  19. monroe17 Group Title
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    \[\frac{ 1 }{ 5 }\int\limits_{?}^{?}(u)^5du\] this..?

    • one year ago
  20. Spacelimbus Group Title
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    perfect

    • one year ago
  21. Spacelimbus Group Title
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    You know ho to integrate that right?

    • one year ago
  22. monroe17 Group Title
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    uhm, i dont, this is where i get stuck the majority of the time. i dont get this part

    • one year ago
  23. Spacelimbus Group Title
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    would you know hot integrate this expression \[ \large \int x^3dx \] ?

    • one year ago
  24. Spacelimbus Group Title
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    excuse my poor spelling (-; how to *

    • one year ago
  25. monroe17 Group Title
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    isn't it just finding the antideriv?

    • one year ago
  26. Spacelimbus Group Title
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    exactly

    • one year ago
  27. Spacelimbus Group Title
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    with the general rule: \[ \Large \int x^ndx = \frac{1}{n+1}x^{n+1}+C\]

    • one year ago
  28. Spacelimbus Group Title
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    You've managed to simplify your integral to exact that form, except that in your case it says \(u\) rather than \(x\)

    • one year ago
  29. monroe17 Group Title
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    okay hold on :)

    • one year ago
  30. monroe17 Group Title
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    \[\frac{ 1 }{ 5 }\int\limits_{?}^{?}\frac{ 1 }{ 6 }u^6+C\]

    • one year ago
  31. Spacelimbus Group Title
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    exactly, now you can just multiply it out and you're done.

    • one year ago
  32. Spacelimbus Group Title
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    well, after backsubstitution you're done \[ u = \ln x \]

    • one year ago
  33. monroe17 Group Title
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    what do you mean multiply it out? the 1/6 ?

    • one year ago
  34. Spacelimbus Group Title
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    by the way, I guess you mean the right thing, but you're spelling is a bit incorrect. You better write it like that: \[ \large \frac{1}{5}\int u^5du = \frac{1}{5}\left( \frac{1}{6}u^6+C\right) \]

    • one year ago
  35. Spacelimbus Group Title
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    see the general rule of integration above again. the RHS doesn't include any integral sign anymore.

    • one year ago
  36. monroe17 Group Title
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    oh i get it!

    • one year ago
  37. Spacelimbus Group Title
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    great (-: !

    • one year ago
  38. monroe17 Group Title
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    Thank you, you're great at explaining :)

    • one year ago
  39. Spacelimbus Group Title
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    Appreciate it, you're welcome

    • one year ago
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