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

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heradog Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}\frac{ \ln(x)^4 }{ x }dx\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
try \(u=ln(x), du =\frac{dx}{x}\) and you should get it in one step
 one year ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.0
May I help?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
visualize it as \[\int\frac{ \ln(x)^4 }{ x }dx=\int \ln^4(x)\frac{dx}{x}\]
 one year ago

heradog Group TitleBest ResponseYou've already chosen the best response.0
pretend I'm a complete moron
 one year ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.0
(x)^4 ?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
are you familiar with "u  substitution"
 one year ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.0
\(\ln x^4\) Why \(\ln ^4 x\) @satellite73
 one year ago

heradog Group TitleBest ResponseYou've already chosen the best response.0
sorry, it's the whole thing to the 4th
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
it is the chain rule backwards
 one year ago

heradog Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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 \(3\ln(x)\times \frac{1}{x}=\frac{3\ln^2(x)}{x}\)
 one year ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.0
Ok! \((\ln x)^4\) is what we have
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
this tells you \[\int\frac{3\ln^2(x)}{x}dx=\ln^3(x)\]
 one year ago

heradog Group TitleBest ResponseYou've already chosen the best response.0
yes @SheldonEinstein
 one year ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.0
I would not interupt and let @satellite73 answer.
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
so a usub 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" \(\frac{1}{x}\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
if you make the substitution \(u=\ln(x)\) then the derivative is \(\frac{1}{x}\) so the gimmick is to say \[u=\ln(x), du=\frac{1}{x}dx\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
then rewrite with only \(u\) terms and get \[\int u^4du\]
 one year ago

heradog Group TitleBest ResponseYou've already chosen the best response.0
Ok I'm with you so far
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
now the anti derivative is relatively easy it is \(\frac{u^5}{5}\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
putting the \(u=\ln(x)\) back we get \[\frac{\ln^5(x)}{5}\]
 one year ago

heradog Group TitleBest ResponseYou've already chosen the best response.0
why over 5? that's where I got lost
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
check that this works by differentiation, and you will see that this is correct
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
\[\int x^n dx =\frac{x^{n+1}}{n+1}\]
 one year ago

heradog Group TitleBest ResponseYou've already chosen the best response.0
I'm used to doing (1/5)u^5
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

heradog Group TitleBest ResponseYou've already chosen the best response.0
both of those really help! I'll try the next one and see what happens
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
the entire problem is to figure out the \[\frac{1}{5}\ln^5(x)\] part
 one year ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.0
Ok! I got it. @heradog not me, @satellite73 was the one who helped. Best of luck for your journey and btw \(\mathbb{ I} \space \mathbb{LOVE} \space \mathbb{INTEGRALS} \)
 one year ago

heradog Group TitleBest ResponseYou've already chosen the best response.0
good i'm stuck again
 one year ago
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