Question

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

Answer 1

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

Answer 2

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

Answer 3

May I help?

Answer 4

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

Answer 5

pretend I'm a complete moron

Answer 6

(x)^4 ?

Answer 7

are you familiar with "u - substitution"

Answer 8

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

Answer 9

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

Answer 10

it is the chain rule backwards

Answer 11

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

Answer 12

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 \(3\ln(x)\times \frac{1}{x}=\frac{3\ln^2(x)}{x}\)

Answer 13

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

Answer 14

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

Answer 15

yes @SheldonEinstein

Answer 16

I would not interupt and let @satellite73 answer.

Answer 17

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" \(\frac{1}{x}\)

Answer 18

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

Answer 19

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

Answer 20

Ok I'm with you so far

Answer 21

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

Answer 22

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

Answer 23

why over 5? that's where I got lost

Answer 24

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

Answer 25

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

Answer 26

nevermind

Answer 27

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

Answer 28

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

Answer 29

ok

Answer 30

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

Answer 31

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 ?

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

Ok! I got it.

@heradog not me, @satellite73 was the one who helped. Best of luck for your journey and btw

\(\mathbb{ I} \space \mathbb{LOVE} \space \mathbb{INTEGRALS} \)

Answer 36

good i'm stuck again