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heradog

  • one year ago

I need help with this step by step Evaluate the indefinite integral. \int \frac{(\ln(x))^4}{x} dx by substitution

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  1. heradog
    • one year ago
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    \[\int\limits_{}^{}\frac{ \ln(x)^4 }{ x }dx\]

  2. satellite73
    • one year ago
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    try \(u=ln(x), du =\frac{dx}{x}\) and you should get it in one step

  3. SheldonEinstein
    • one year ago
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    May I help?

  4. satellite73
    • one year ago
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    visualize it as \[\int\frac{ \ln(x)^4 }{ x }dx=\int \ln^4(x)\frac{dx}{x}\]

  5. heradog
    • one year ago
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    pretend I'm a complete moron

  6. SheldonEinstein
    • one year ago
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    (x)^4 ?

  7. satellite73
    • one year ago
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    are you familiar with "u - substitution"

  8. SheldonEinstein
    • one year ago
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    \(\ln x^4\) Why \(\ln ^4 x\) @satellite73

  9. heradog
    • one year ago
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    sorry, it's the whole thing to the 4th

  10. satellite73
    • one year ago
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    it is the chain rule backwards

  11. heradog
    • one year ago
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    I understand that you pick one component to make u and that the derivative of that needs to be in the function it's what you do with it that I can't figure out

  12. satellite73
    • one year ago
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    first off, lets get the answer and see why it works. what do you get when you take the derivative of \(\ln^3(x)\) ? by the chain rule you get \(3\ln(x)\times \frac{1}{x}=\frac{3\ln^2(x)}{x}\)

  13. SheldonEinstein
    • one year ago
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    Ok! \((\ln x)^4\) is what we have

  14. satellite73
    • one year ago
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    this tells you \[\int\frac{3\ln^2(x)}{x}dx=\ln^3(x)\]

  15. heradog
    • one year ago
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    yes @SheldonEinstein

  16. SheldonEinstein
    • one year ago
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    I would not interupt and let @satellite73 answer.

  17. satellite73
    • one year ago
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    so a u-sub is a way to undo the chain rule you have a composite function \(\ln^4(x)=(\ln(x))^4\) multiplied by the derivative of the "inside function" \(\frac{1}{x}\)

  18. satellite73
    • one year ago
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    if you make the substitution \(u=\ln(x)\) then the derivative is \(\frac{1}{x}\) so the gimmick is to say \[u=\ln(x), du=\frac{1}{x}dx\]

  19. satellite73
    • one year ago
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    then rewrite with only \(u\) terms and get \[\int u^4du\]

  20. heradog
    • one year ago
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    Ok I'm with you so far

  21. satellite73
    • one year ago
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    now the anti derivative is relatively easy it is \(\frac{u^5}{5}\)

  22. satellite73
    • one year ago
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    putting the \(u=\ln(x)\) back we get \[\frac{\ln^5(x)}{5}\]

  23. heradog
    • one year ago
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    why over 5? that's where I got lost

  24. satellite73
    • one year ago
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    check that this works by differentiation, and you will see that this is correct

  25. satellite73
    • one year ago
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    oh, over 5 because the derivative of \(u^5\) is \(5u^4\) but you want \(u^4\) and not \(5u^4\) so you have to divide by \(5\) to get it

  26. heradog
    • one year ago
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    nevermind

  27. satellite73
    • one year ago
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    \[\int x^n dx =\frac{x^{n+1}}{n+1}\]

  28. heradog
    • one year ago
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    I'm used to doing (1/5)u^5

  29. satellite73
    • one year ago
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    ok

  30. satellite73
    • one year ago
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    check that the derivative of \[\frac{1}{5}\ln^5(x)\] is \[\frac{\ln^4(x)}{x}\] and you will see why the "u - substitution" works

  31. SheldonEinstein
    • one year ago
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    I did like this : \[\int \cfrac{ [\ln x]^4 dx }{x} \] Put \(\cfrac{1}{5} [\ln x]^5 = u \) \(\int du = u + c\) = \(\cfrac{1}{5} [\ln x]^5 + C\) \(\implies \cfrac{[\ln x]^4 dx}{x} = du\) Is this also correct @satellite73 ?

  32. satellite73
    • one year ago
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    it looks like you knew the answer at the start and then just said it look at this line \[\int du = u + c\] this says the integral is \(u\) which you knew at the beginning, probably by doing the u - sub in your head in other words, you knew what the answer was, but the u - sub is the way to get the answer

  33. heradog
    • one year ago
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    both of those really help! I'll try the next one and see what happens

  34. satellite73
    • one year ago
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    the entire problem is to figure out the \[\frac{1}{5}\ln^5(x)\] part

  35. SheldonEinstein
    • one year ago
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    Ok! I got it. @heradog not me, @satellite73 was the one who helped. Best of luck for your journey and btw \(\mathbb{ I} \space \mathbb{LOVE} \space \mathbb{INTEGRALS} \)

  36. heradog
    • one year ago
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    good i'm stuck again

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