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monroe17

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

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

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

  8. Spacelimbus
    • one year 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
    • one year ago
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    did it come from the 1/5x part?

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

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

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

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

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

  15. Spacelimbus
    • one year 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
    • one year ago
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    oh okayy.. so now do i find the anti derivative?

  17. Spacelimbus
    • one year 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
    • one year ago
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    lol! okay :) give me a second

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

  20. Spacelimbus
    • one year ago
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    perfect

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

  22. monroe17
    • one year 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
    • one year ago
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    would you know hot integrate this expression \[ \large \int x^3dx \] ?

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

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

  26. Spacelimbus
    • one year ago
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    exactly

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

  28. Spacelimbus
    • one year 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
    • one year ago
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    okay hold on :)

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

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

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

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

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

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

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

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

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

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