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

- anonymous

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

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- anonymous

start by factoring

- anonymous

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

- 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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## More answers

- anonymous

@Elsa213 i am pretty sure that is some other answer to some other question

- Elsa213

ik ik

- Elsa213

it was just an example

- anonymous

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

- xapproachesinfinity

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

- anonymous

right

- anonymous

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

- xapproachesinfinity

you forgot log lol

- anonymous

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

- anonymous

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

- 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

yes continue this way

- anonymous

ok thanks @xapproachesinfinity and @satellite73

- xapproachesinfinity

you should know what is int log|x|dx

- 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

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

- anonymous

i would say this integral looks nothing like that at all

- xapproachesinfinity

satellite you assumed that the expression will take positive values only
you need the absolute value theere

- xapproachesinfinity

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

- anonymous

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

- anonymous

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

- xapproachesinfinity

yes i doesn't :)

- xapproachesinfinity

it*

- anonymous

or dx/x^2-a^2

- 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

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

- anonymous

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

- anonymous

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

- anonymous

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

- xapproachesinfinity

but you excluded the negative values?

- ganeshie8

why wolfram is giving this mess
http://www.wolframalpha.com/input/?i=%5Cint+log%28abs%284x%5E2-1%29%29

- xapproachesinfinity

wolfram is crazy hehe

- xapproachesinfinity

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

- 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

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

- myininaya

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

- myininaya

integration by parts will be needed
polynomial division
partial fractions look like they are needed as well

- xapproachesinfinity

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

- xapproachesinfinity

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

- xapproachesinfinity

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

- 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

so far it looks nice to me

- myininaya

where did you get that @xapproachesinfinity ?

- xapproachesinfinity

|dw:1408559618618:dw|

- xapproachesinfinity

from by part integration

- xapproachesinfinity

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

- myininaya

I don't see where that integral comes from

- myininaya

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

- xapproachesinfinity

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

- myininaya

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

- xapproachesinfinity

yup you are right forgot that xD

- xapproachesinfinity

it should be what you wrote above

- myininaya

@cutelilgirl are you still there?

- xapproachesinfinity

8x^2/4x^2-1

- anonymous

yes but this seems a little beyond me :(

- myininaya

have you talked about integration by parts?

- myininaya

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

- myininaya

so i assume you have had it mentioned in class

- myininaya

is that a correct assumption

- anonymous

yes

- xapproachesinfinity

can't do partial fraction here

- myininaya

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

- myininaya

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

- myininaya

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

- xapproachesinfinity

we need another integration by part? i guess

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