anonymous
  • anonymous
how to integrate log | 4x^2-1| dx ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
start by factoring
anonymous
  • anonymous
then break apart in to two pieces then integrate term by term
Elsa213
  • Elsa213
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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anonymous
  • anonymous
@Elsa213 i am pretty sure that is some other answer to some other question
Elsa213
  • Elsa213
ik ik
Elsa213
  • Elsa213
it was just an example
anonymous
  • anonymous
log | (2x+1)(2x-1) | ??
xapproachesinfinity
  • xapproachesinfinity
log a.b=loga +logb then integrate each at a time
anonymous
  • anonymous
right
anonymous
  • anonymous
now \[\int \log(2x+1)dx+\int (2x-1)dx\]
xapproachesinfinity
  • xapproachesinfinity
you forgot log lol
anonymous
  • anonymous
then you should be done because you should have the integral of the log memorized
anonymous
  • anonymous
so i did \[\int \log(2x+1)dx+\int \log(2x-1)dx\]
anonymous
  • anonymous
the answer given is : x log | 4x^2-1| - 1/2 log | (2x-1)/(2x+1) | .. will i get this answer this way ?
xapproachesinfinity
  • xapproachesinfinity
yes continue this way
anonymous
  • anonymous
ok thanks @xapproachesinfinity and @satellite73
xapproachesinfinity
  • xapproachesinfinity
you should know what is int log|x|dx
anonymous
  • anonymous
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
anonymous
  • anonymous
because this exercise was about the problems of integral dx / (a^2_x^2) type
anonymous
  • anonymous
i would say this integral looks nothing like that at all
xapproachesinfinity
  • xapproachesinfinity
satellite you assumed that the expression will take positive values only you need the absolute value theere
xapproachesinfinity
  • xapproachesinfinity
i would say this integral looks nothing like that at all === is this to my reply^_^
anonymous
  • anonymous
yes the | 4x^2 -1 | was there
anonymous
  • anonymous
i would say \[\int \log(4x^2-1)dx\] looks nothing like \[\int \frac{dx}{a^2-x^2}\]
xapproachesinfinity
  • xapproachesinfinity
yes i doesn't :)
xapproachesinfinity
  • xapproachesinfinity
it*
anonymous
  • anonymous
or dx/x^2-a^2
anonymous
  • anonymous
\[\int \log(x)dx=x\log(x)-x\] 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
xapproachesinfinity
  • xapproachesinfinity
@satellite73 , any reason why you took off absolute values?
anonymous
  • anonymous
damn typo \[\int \log(2x+1)dx=\frac{1}{2}\left((2x+1)\log(2x+1)-(2x+1)\right)\]
anonymous
  • anonymous
or if you prefer \[\frac{1}{2}(2x+1)(\log(2x+1)-1)\]
anonymous
  • anonymous
laziness is all that and the stupid \(+ c\) got omitted as well
xapproachesinfinity
  • xapproachesinfinity
but you excluded the negative values?
ganeshie8
  • ganeshie8
why wolfram is giving this mess http://www.wolframalpha.com/input/?i=%5Cint+log%28abs%284x%5E2-1%29%29
xapproachesinfinity
  • xapproachesinfinity
wolfram is crazy hehe
xapproachesinfinity
  • xapproachesinfinity
i say the absolute value cannot be taken off. as we are excluding negative interval here
anonymous
  • anonymous
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
xapproachesinfinity
  • xapproachesinfinity
\(\Large \rm\color{midnightblue}{\int log|(2x+1)(2x-1)|dx=\int log|2x+1|dx+\\ \int log|2x-1|dx}\)
myininaya
  • myininaya
Integration by parts should even work for the first way this expression appears
myininaya
  • myininaya
integration by parts will be needed polynomial division partial fractions look like they are needed as well
xapproachesinfinity
  • xapproachesinfinity
i agree. i looks that the answer would look much simpler
xapproachesinfinity
  • xapproachesinfinity
x/(2x-1)(2x+1) partial fraction here
xapproachesinfinity
  • xapproachesinfinity
the other part is good xlog|4x^2-1|
myininaya
  • myininaya
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) \[=x \ln(f(x))-\int\limits_{}^{}x \frac{f'(x)}{f(x)} dx\]
xapproachesinfinity
  • xapproachesinfinity
so far it looks nice to me
myininaya
  • myininaya
where did you get that @xapproachesinfinity ?
xapproachesinfinity
  • xapproachesinfinity
|dw:1408559618618:dw|
xapproachesinfinity
  • xapproachesinfinity
from by part integration
xapproachesinfinity
  • xapproachesinfinity
this now going to that expression she wrote as the answer!
myininaya
  • myininaya
I don't see where that integral comes from
myininaya
  • myininaya
i wonder if you forget f' on top |dw:1408559852906:dw|
xapproachesinfinity
  • xapproachesinfinity
u=log(|4x^2-1|) ===> du=1/4x^2-1 dv=1 ====> v=x may be there is an error lol
myininaya
  • myininaya
derivative of ln(f) is equal to f'/f
xapproachesinfinity
  • xapproachesinfinity
yup you are right forgot that xD
xapproachesinfinity
  • xapproachesinfinity
it should be what you wrote above
myininaya
  • myininaya
@cutelilgirl are you still there?
xapproachesinfinity
  • xapproachesinfinity
8x^2/4x^2-1
anonymous
  • anonymous
yes but this seems a little beyond me :(
myininaya
  • myininaya
have you talked about integration by parts?
myininaya
  • myininaya
that is the only way i know how to do this problem
myininaya
  • myininaya
so i assume you have had it mentioned in class
myininaya
  • myininaya
is that a correct assumption
anonymous
  • anonymous
yes
xapproachesinfinity
  • xapproachesinfinity
can't do partial fraction here
myininaya
  • myininaya
do you know how to evaluate \[\int \ln(x) dx ]
myininaya
  • myininaya
\[\int\limits_{}^{} \ln(x) dx\]
myininaya
  • myininaya
If you don't I can show you that may help you seeing this more simple integral evaluated first
xapproachesinfinity
  • xapproachesinfinity
we need another integration by part? i guess

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