Question

Now What?

Answer 1

I'm a hedgehog

Answer 2

On a less serious note, you might want to post a question.

Answer 3

\[\int\limits_{}^{}\frac{ 8dx }{ xln(3x) }\]

Answer 4

was working on it....so it's integration by substitution but I don't know what to do with the dx inside

Answer 5

\(\large \int\frac{dx}{x \ln x}\)
take u=ln x
then du=dx/x
=>\(\large \int\frac{du}{u}=\ln u =>\ln(\ln x)+C\)

Answer 6

I'm just putting this classic problem out because yours is a variant. I"m not going to tell you what u should be, cause imo that takes out the fun.

Answer 7

tell me if you need more detail...

Answer 8

substitute \[\ln(3x)=u\implies {3\over3x}dx=du\]

Answer 9

I understand the u and du but how do I go back to the original function with it?

Answer 10

well take the final expression in u and resubstitute whatever you took u to be in the first place for x, so there are only x's in the expression (or t's, or whatevers)

Answer 11

your I will look of this sort now
\[
I=8\int\frac{1}{\ln(3x)}\frac{dx}{x}=8\int{1\over u}du
\]

Answer 12

this is what you'd be integrating now

Answer 13

you follow @heradog ?

Answer 14

Yeah that makes a lot more sense

Answer 15

but I"m still not getting the final answer right

Answer 16

show your steps here so ..

Answer 17

\[8\int\limits_{}^{}\frac{ 1 }{ u }du\]
\[8*\frac{ 1 }{ \ln (3x) } +C\]

Answer 18

haha
you skipped a step
integrate with "u" not "x"

Answer 19

antiderivative should be 8*ln(u) +C

Answer 20

\[\int{du\over u}=?\]

Answer 21

forget the relation between "u" and "x" for that step

Answer 22

so is 8ln(u)+C even close or am I way off?

Answer 23

you are back on track now

Answer 24

so what do I do from there? nothing works

Answer 25

now you can happily replace your "u" with "x"
u = ln(3x)

Answer 26

that "u" is like a bag
pack your stuff in it, you take the bag with you, you get you stuff. when you are done carrying, get rid of it!

Answer 27

but that doesn't work
8ln(ln(3x)+C says its wrong

Answer 28

\[{d\over dx}8\ln(\ln(3x))+C=8{1\over\ln(3x)}{1\over3x}(3)=\frac{8}{x\ln(3x)}
\]

check: \[\int\frac{8}{x\ln(3x)}dx\]

Answer 29

none of that works either

Answer 30

did you check the problem itself?

Answer 31

coz, for the problem given, the antiderivative is correct as QED by the derivative itself.

Answer 32

ok, do this

Answer 33

@heradog
\[I=8\ln|\ln(3x)|+C\]
careful with the modulus sign

Answer 34

@heradog hello?

Answer 35

thanks

Answer 36

works, right?

Answer 37

yeah it did

Answer 38

now, to commemorate this, a medal is in order

Answer 39

<:)

Answer 40

I think 5 is better :)

Answer 41

okey hi how are u doing i have this guy that will really help u get the concept if u want

Answer 42

definitely can't hurt me any @jack63

Answer 43

okey

Answer 44

iam bringing him in now

Answer 45

he is coming

Answer 46

so is this your first question

Answer 47

no it is not

Answer 48

i will be right back stay right where okey

Answer 49

I've been working on different ones all night
I got the work done enough for tonight but will need help on the next one

Answer 50

were do u live

Answer 51

mountain time

Answer 52

cool in america

Answer 53

yeah

Answer 54

cool i am living somewhere by russia

Answer 55

there he is he is mathslover he will help u

Answer 56

thanks

Answer 57

Didn't @electrokid , @inkyvoyd explained already?

Answer 58

yeah I got it finally

Answer 59

That's nice.

Answer 60

@electrokid has been really good at walking me through it all

Answer 61

heradog if u need to watch some videos to help u go to teachertube

Answer 62

He is really great.

Answer 63

jack63 offered you as extra help and I didn't want to turn it down, the work for tonight is already closed, but I'm sure I'll need help on the next one which is integration by parts and will probably start on it in the next couple days

Answer 64

best of luck for that work!

Answer 65

thanks

Answer 66

good luck heradog

Answer 67

thanks jack

Answer 68

@heradog you are welcome