Question

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

Answer 1

start by factoring

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

ik ik

Answer 6

it was just an example

Answer 7

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

Answer 8

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

Answer 9

right

Answer 10

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

Answer 11

you forgot log lol

Answer 12

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

Answer 13

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

Answer 14

the answer given is : x log | 4x^2-1| - 1/2 log | (2x-1)/(2x+1) | .. will i get this answer this way ?

Answer 15

yes continue this way

Answer 16

ok thanks @xapproachesinfinity and @satellite73

Answer 17

you should know what is int log|x|dx

Answer 18

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

Answer 19

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

Answer 20

i would say this integral looks nothing like that at all

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

yes i doesn't :)

Answer 26

it*

Answer 27

or dx/x^2-a^2

Answer 28

\[\int \log(x)dx=x\log(x)-x\] 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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

but you excluded the negative values?

Answer 34

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

Answer 35

wolfram is crazy hehe

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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)
\[=x \ln(f(x))-\int\limits_{}^{}x \frac{f'(x)}{f(x)} dx\]

Answer 45

so far it looks nice to me

Answer 46

where did you get that @xapproachesinfinity ?