anonymous
  • anonymous
I need help with this step by step Evaluate the indefinite integral. \int \frac{(\ln(x))^4}{x} dx by substitution
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits_{}^{}\frac{ \ln(x)^4 }{ x }dx\]
anonymous
  • anonymous
try \(u=ln(x), du =\frac{dx}{x}\) and you should get it in one step
anonymous
  • anonymous
May I help?

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anonymous
  • anonymous
visualize it as \[\int\frac{ \ln(x)^4 }{ x }dx=\int \ln^4(x)\frac{dx}{x}\]
anonymous
  • anonymous
pretend I'm a complete moron
anonymous
  • anonymous
(x)^4 ?
anonymous
  • anonymous
are you familiar with "u - substitution"
anonymous
  • anonymous
\(\ln x^4\) Why \(\ln ^4 x\) @satellite73
anonymous
  • anonymous
sorry, it's the whole thing to the 4th
anonymous
  • anonymous
it is the chain rule backwards
anonymous
  • anonymous
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
anonymous
  • anonymous
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 \(3\ln(x)\times \frac{1}{x}=\frac{3\ln^2(x)}{x}\)
anonymous
  • anonymous
Ok! \((\ln x)^4\) is what we have
anonymous
  • anonymous
this tells you \[\int\frac{3\ln^2(x)}{x}dx=\ln^3(x)\]
anonymous
  • anonymous
yes @SheldonEinstein
anonymous
  • anonymous
I would not interupt and let @satellite73 answer.
anonymous
  • anonymous
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" \(\frac{1}{x}\)
anonymous
  • anonymous
if you make the substitution \(u=\ln(x)\) then the derivative is \(\frac{1}{x}\) so the gimmick is to say \[u=\ln(x), du=\frac{1}{x}dx\]
anonymous
  • anonymous
then rewrite with only \(u\) terms and get \[\int u^4du\]
anonymous
  • anonymous
Ok I'm with you so far
anonymous
  • anonymous
now the anti derivative is relatively easy it is \(\frac{u^5}{5}\)
anonymous
  • anonymous
putting the \(u=\ln(x)\) back we get \[\frac{\ln^5(x)}{5}\]
anonymous
  • anonymous
why over 5? that's where I got lost
anonymous
  • anonymous
check that this works by differentiation, and you will see that this is correct
anonymous
  • anonymous
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
anonymous
  • anonymous
nevermind
anonymous
  • anonymous
\[\int x^n dx =\frac{x^{n+1}}{n+1}\]
anonymous
  • anonymous
I'm used to doing (1/5)u^5
anonymous
  • anonymous
ok
anonymous
  • anonymous
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
anonymous
  • anonymous
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 ?
anonymous
  • anonymous
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
anonymous
  • anonymous
both of those really help! I'll try the next one and see what happens
anonymous
  • anonymous
the entire problem is to figure out the \[\frac{1}{5}\ln^5(x)\] part
anonymous
  • anonymous
Ok! I got it. @heradog not me, @satellite73 was the one who helped. Best of luck for your journey and btw \(\mathbb{ I} \space \mathbb{LOVE} \space \mathbb{INTEGRALS} \)
anonymous
  • anonymous
good i'm stuck again

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