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monroe17

  • 2 years ago

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

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  1. Spacelimbus
    • 2 years ago
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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?

  2. monroe17
    • 2 years ago
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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 ?

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

  4. Spacelimbus
    • 2 years ago
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    Yes, perfect @monroe17 !! You're pretty close from there.

  5. monroe17
    • 2 years ago
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    wait why and how did you rewrite it like that?

  6. Spacelimbus
    • 2 years ago
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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) \[ \large \frac{(\ln x)^5}{5x}= (\ln x)^5 \cdot \frac{1}{5x} \]

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

  8. Spacelimbus
    • 2 years ago
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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.

  9. monroe17
    • 2 years ago
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    did it come from the 1/5x part?

  10. Spacelimbus
    • 2 years ago
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    exactly, I just thought the integral would look nicer without the unnecessary 5 in the denominator.

  11. monroe17
    • 2 years ago
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    but then why did it stay 1/x ... why wouldn't it be x?

  12. Spacelimbus
    • 2 years ago
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    as an example:\[\Large \int \frac{1}{5x}dx = \frac{1}{5} \int \frac{1}{x}dx \]

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

  14. monroe17
    • 2 years ago
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    oh lol.. okay

  15. Spacelimbus
    • 2 years ago
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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 (-:

  16. monroe17
    • 2 years ago
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    oh okayy.. so now do i find the anti derivative?

  17. Spacelimbus
    • 2 years ago
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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.

  18. monroe17
    • 2 years ago
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    lol! okay :) give me a second

  19. monroe17
    • 2 years ago
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    \[\frac{ 1 }{ 5 }\int\limits_{?}^{?}(u)^5du\] this..?

  20. Spacelimbus
    • 2 years ago
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    perfect

  21. Spacelimbus
    • 2 years ago
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    You know ho to integrate that right?

  22. monroe17
    • 2 years ago
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    uhm, i dont, this is where i get stuck the majority of the time. i dont get this part

  23. Spacelimbus
    • 2 years ago
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    would you know hot integrate this expression \[ \large \int x^3dx \] ?

  24. Spacelimbus
    • 2 years ago
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    excuse my poor spelling (-; how to *

  25. monroe17
    • 2 years ago
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    isn't it just finding the antideriv?

  26. Spacelimbus
    • 2 years ago
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    exactly

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

  28. Spacelimbus
    • 2 years ago
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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 \(x\)

  29. monroe17
    • 2 years ago
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    okay hold on :)

  30. monroe17
    • 2 years ago
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    \[\frac{ 1 }{ 5 }\int\limits_{?}^{?}\frac{ 1 }{ 6 }u^6+C\]

  31. Spacelimbus
    • 2 years ago
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    exactly, now you can just multiply it out and you're done.

  32. Spacelimbus
    • 2 years ago
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    well, after backsubstitution you're done \[ u = \ln x \]

  33. monroe17
    • 2 years ago
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    what do you mean multiply it out? the 1/6 ?

  34. Spacelimbus
    • 2 years ago
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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: \[ \large \frac{1}{5}\int u^5du = \frac{1}{5}\left( \frac{1}{6}u^6+C\right) \]

  35. Spacelimbus
    • 2 years ago
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    see the general rule of integration above again. the RHS doesn't include any integral sign anymore.

  36. monroe17
    • 2 years ago
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    oh i get it!

  37. Spacelimbus
    • 2 years ago
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    great (-: !

  38. monroe17
    • 2 years ago
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    Thank you, you're great at explaining :)

  39. Spacelimbus
    • 2 years ago
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    Appreciate it, you're welcome

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