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start by factoring

then break apart in to two pieces
then integrate term by term

ik ik

it was just an example

log | (2x+1)(2x-1) | ??

log a.b=loga +logb
then integrate each at a time

right

now
\[\int \log(2x+1)dx+\int (2x-1)dx\]

you forgot log lol

then you should be done because you should have the integral of the log memorized

so i did \[\int \log(2x+1)dx+\int \log(2x-1)dx\]

yes continue this way

ok thanks @xapproachesinfinity and @satellite73

you should know what is int log|x|dx

because this exercise was about the problems of integral dx / (a^2_x^2) type

i would say this integral looks nothing like that at all

i would say this integral looks nothing like that at all
===
is this to my reply^_^

yes the | 4x^2 -1 | was there

i would say
\[\int \log(4x^2-1)dx\] looks nothing like
\[\int \frac{dx}{a^2-x^2}\]

yes i doesn't :)

or dx/x^2-a^2

@satellite73 , any reason why you took off absolute values?

damn typo \[\int \log(2x+1)dx=\frac{1}{2}\left((2x+1)\log(2x+1)-(2x+1)\right)\]

or if you prefer
\[\frac{1}{2}(2x+1)(\log(2x+1)-1)\]

laziness is all
that and the stupid \(+
c\) got omitted as well

but you excluded the negative values?

wolfram is crazy hehe

i say the absolute value cannot be taken off. as we are excluding negative interval here

\(\Large \rm\color{midnightblue}{\int log|(2x+1)(2x-1)|dx=\int log|2x+1|dx+\\ \int log|2x-1|dx}\)

Integration by parts should even work for the first way this expression appears

i agree. i looks that the answer would look much simpler

x/(2x-1)(2x+1) partial fraction here

the other part is good xlog|4x^2-1|

so far it looks nice to me

where did you get that @xapproachesinfinity ?

|dw:1408559618618:dw|

from by part integration

this now going to that expression she wrote as the answer!

I don't see where that integral comes from

i wonder if you forget f' on top
|dw:1408559852906:dw|

u=log(|4x^2-1|) ===> du=1/4x^2-1
dv=1 ====> v=x
may be there is an error lol

derivative of ln(f) is equal to f'/f

yup you are right forgot that xD

it should be what you wrote above

@cutelilgirl are you still there?

8x^2/4x^2-1

yes but this seems a little beyond me :(

have you talked about integration by parts?

that is the only way i know how to do this problem

so i assume you have had it mentioned in class

is that a correct assumption

yes

can't do partial fraction here

do you know how to evaluate
\[\int \ln(x) dx ]

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

If you don't I can show you
that may help you seeing this more simple integral evaluated first

we need another integration by part? i guess