how to integrate log | 4x^2-1| dx ?

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how to integrate log | 4x^2-1| dx ?

Mathematics
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start by factoring
then break apart in to two pieces then integrate term by term
erm.... integral 4x/(1+x^2) = I Let 1+x^2 = t ...........(1) --> 2x = dt/dx --> 2x dx = dt ...........(2) Now I = integral 2 [2x dx / (1+x^2) ] = integral 2 [dt / t ] since from (2) = 2 [integral dt / t ] = 2 (ln t) +c = 2 ln(1+x^2) + c We have integral 1/t dt = ln t +c Therefore I = 2 ln(1+x^2) +c where ln is log to the base e https://answers.yahoo.com/question/index?qid=20110416060345AAVswO7

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@Elsa213 i am pretty sure that is some other answer to some other question
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 \[\int \log(2x+1)dx+\int \log(2x-1)dx\]
the answer given is : x log | 4x^2-1| - 1/2 log | (2x-1)/(2x+1) | .. will i get this answer this way ?
yes continue this way
ok thanks @xapproachesinfinity and @satellite73
you should know what is int log|x|dx
i have no idea when you integrate you will get something then using the properties of the log you can change it in to many different forms don't get married to the answer in the book
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
satellite you assumed that the expression will take positive values only you need the absolute value theere
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 :)
it*
or dx/x^2-a^2
\[\int \log(x)dx=x\log(x)-x\] so \[\int \log(2x+1)dx=\frac{1}{2}\left((2x+1)\log(2x-1)-(2x+1)\right)\] by a mental u sub second part is similar
@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?
why wolfram is giving this mess http://www.wolframalpha.com/input/?i=%5Cint+log%28abs%284x%5E2-1%29%29
wolfram is crazy hehe
i say the absolute value cannot be taken off. as we are excluding negative interval here
wolfram gave that mess because of the sign problem here it is neater http://www.wolframalpha.com/input/?i=\int+log%284x^2-1%29
\(\Large \rm\color{midnightblue}{\int log|(2x+1)(2x-1)|dx=\int log|2x+1|dx+\\ \int log|2x-1|dx}\)
Integration by parts should even work for the first way this expression appears
integration by parts will be needed polynomial division partial fractions look like they are needed as well
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|
Whenever I see something that looks like \[\int\limits_{}^{}1 \cdot \ln(f(x)) dx \] one of my first instincts it to go with the integration by parts (assuming f>0) \[=x \ln(f(x))-\int\limits_{}^{}x \frac{f'(x)}{f(x)} dx\]
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

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