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

- anonymous

- jamiebookeater

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

\[\int\limits_{}^{}\frac{ \ln(x)^4 }{ x }dx\]

- anonymous

try \(u=ln(x), du =\frac{dx}{x}\) and you should get it in one step

- anonymous

May I help?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

visualize it as
\[\int\frac{ \ln(x)^4 }{ x }dx=\int \ln^4(x)\frac{dx}{x}\]

- anonymous

pretend I'm a complete moron

- anonymous

(x)^4 ?

- anonymous

are you familiar with "u - substitution"

- anonymous

\(\ln x^4\) Why \(\ln ^4 x\) @satellite73

- anonymous

sorry, it's the whole thing to the 4th

- anonymous

it is the chain rule backwards

- 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

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

Ok! \((\ln x)^4\) is what we have

- anonymous

this tells you
\[\int\frac{3\ln^2(x)}{x}dx=\ln^3(x)\]

- anonymous

yes @SheldonEinstein

- anonymous

I would not interupt and let @satellite73 answer.

- 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

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

then rewrite with only \(u\) terms and get
\[\int u^4du\]

- anonymous

Ok I'm with you so far

- anonymous

now the anti derivative is relatively easy
it is \(\frac{u^5}{5}\)

- anonymous

putting the \(u=\ln(x)\) back we get
\[\frac{\ln^5(x)}{5}\]

- anonymous

why over 5? that's where I got lost

- anonymous

check that this works by differentiation, and you will see that this is correct

- 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

nevermind

- anonymous

\[\int x^n dx =\frac{x^{n+1}}{n+1}\]

- anonymous

I'm used to doing (1/5)u^5

- anonymous

ok

- 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

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

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

both of those really help! I'll try the next one and see what happens

- anonymous

the entire problem is to figure out the
\[\frac{1}{5}\ln^5(x)\] part

- 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

good i'm stuck again

Looking for something else?

Not the answer you are looking for? Search for more explanations.