## cutelilgirl one year ago how to integrate log | 4x^2-1| dx ?

1. satellite73

start by factoring

2. satellite73

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

3. 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

4. satellite73

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

5. Elsa213

ik ik

6. Elsa213

it was just an example

7. cutelilgirl

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

8. xapproachesinfinity

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

9. satellite73

right

10. satellite73

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

11. xapproachesinfinity

you forgot log lol

12. satellite73

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

13. satellite73

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

14. cutelilgirl

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

15. xapproachesinfinity

yes continue this way

16. cutelilgirl

ok thanks @xapproachesinfinity and @satellite73

17. xapproachesinfinity

you should know what is int log|x|dx

18. satellite73

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

19. cutelilgirl

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

20. satellite73

i would say this integral looks nothing like that at all

21. xapproachesinfinity

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

22. xapproachesinfinity

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

23. cutelilgirl

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

24. satellite73

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

25. xapproachesinfinity

yes i doesn't :)

26. xapproachesinfinity

it*

27. cutelilgirl

or dx/x^2-a^2

28. satellite73

$\int \log(x)dx=x\log(x)-x$ 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

29. xapproachesinfinity

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

30. satellite73

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

31. satellite73

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

32. satellite73

laziness is all that and the stupid $$+ c$$ got omitted as well

33. xapproachesinfinity

but you excluded the negative values?

34. ganeshie8

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

35. xapproachesinfinity

wolfram is crazy hehe

36. xapproachesinfinity

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

37. satellite73

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

38. xapproachesinfinity

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

39. myininaya

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

40. myininaya

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

41. xapproachesinfinity

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

42. xapproachesinfinity

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

43. xapproachesinfinity

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

44. 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) $=x \ln(f(x))-\int\limits_{}^{}x \frac{f'(x)}{f(x)} dx$

45. xapproachesinfinity

so far it looks nice to me

46. myininaya

where did you get that @xapproachesinfinity ?

47. xapproachesinfinity

|dw:1408559618618:dw|

48. xapproachesinfinity

from by part integration

49. xapproachesinfinity

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

50. myininaya

I don't see where that integral comes from

51. myininaya

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

52. xapproachesinfinity

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

53. myininaya

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

54. xapproachesinfinity

yup you are right forgot that xD

55. xapproachesinfinity

it should be what you wrote above

56. myininaya

@cutelilgirl are you still there?

57. xapproachesinfinity

8x^2/4x^2-1

58. cutelilgirl

yes but this seems a little beyond me :(

59. myininaya

have you talked about integration by parts?

60. myininaya

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

61. myininaya

so i assume you have had it mentioned in class

62. myininaya

is that a correct assumption

63. cutelilgirl

yes

64. xapproachesinfinity

can't do partial fraction here

65. myininaya

do you know how to evaluate $\int \ln(x) dx ] 66. myininaya \[\int\limits_{}^{} \ln(x) dx$

67. myininaya

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

68. xapproachesinfinity

we need another integration by part? i guess