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anilorap

  • 4 years ago

i want to confirm the answer of... integrate lnx/x dx

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  1. anilorap
    • 4 years ago
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    my answer is ln lxl +c

  2. myininaya
    • 4 years ago
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    let u=lnx du=dx/x \[\int\limits_{}^{}u du=\frac{u^2}{2}+C=\frac{(lnx)^2}{2}+C\] we can even check this: \[(\frac{(lnx)^2}{2}+C)'=2*\frac{lnx}{2}*\frac{1}{x}+0=\frac{lnx}{x}\]

  3. myininaya
    • 4 years ago
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    do you got it?

  4. anilorap
    • 4 years ago
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    yes

  5. myininaya
    • 4 years ago
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    k :)

  6. anilorap
    • 4 years ago
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    q. how do i know what to choose?

  7. myininaya
    • 4 years ago
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    i was looking for \[\int\limits_{}^{}f(x)*f'(x)dx\]

  8. myininaya
    • 4 years ago
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    sometimes its not always that easy though

  9. myininaya
    • 4 years ago
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    someone said it is like an art and i agree

  10. anilorap
    • 4 years ago
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    jajaj..ok

  11. myininaya
    • 4 years ago
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    it is not always easy to know what substititon to make but since it was in the form of above it was easy let u=f(x) du=f'(x) dx

  12. myininaya
    • 4 years ago
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    \[\int\limits_{}^{}\frac{lnx}{x}dx=\int\limits_{}^{}lnx*\frac{1}{x} dx\]

  13. myininaya
    • 4 years ago
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    so it was easy for me to notice pretty fast that (lnx)'=1/x so thats why i let u=lnx

  14. anilorap
    • 4 years ago
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    ok..perfect

  15. anilorap
    • 4 years ago
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    so how can i integrate tanx dx..... even though i know thats it is equal to -lnlcosxl+c

  16. myininaya
    • 4 years ago
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    \[\int\limits_{}^{}\frac{sinx}{cosx} dx\] right?

  17. myininaya
    • 4 years ago
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    tanx=sinx/cosx agree?

  18. anilorap
    • 4 years ago
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    ohhhhhhhh... ok

  19. anilorap
    • 4 years ago
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    yes

  20. anilorap
    • 4 years ago
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    so that du/u

  21. myininaya
    • 4 years ago
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    now what do you think the substitution will be?

  22. anilorap
    • 4 years ago
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    with the negative sign of the sin

  23. anilorap
    • 4 years ago
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    u=cos

  24. anilorap
    • 4 years ago
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    du=sin x

  25. anilorap
    • 4 years ago
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    sorry du=-sinx

  26. myininaya
    • 4 years ago
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    du=-sinx dx (i like not to forget about this dx part)

  27. anilorap
    • 4 years ago
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    oh yea

  28. myininaya
    • 4 years ago
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    so show me what that gives us don't integrate yet

  29. anilorap
    • 4 years ago
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    then u have integral of du/u

  30. myininaya
    • 4 years ago
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    negative or positive?

  31. anilorap
    • 4 years ago
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    - integral of du/u

  32. myininaya
    • 4 years ago
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    yes and now integrate

  33. anilorap
    • 4 years ago
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    so my result will be -ln/u/+c

  34. anilorap
    • 4 years ago
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    -ln/cosx/+c

  35. myininaya
    • 4 years ago
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    yes :)

  36. anilorap
    • 4 years ago
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    yeaaa thank uuuu

  37. myininaya
    • 4 years ago
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    np

  38. myininaya
    • 4 years ago
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    so whats this \[\int\limits_{}^{}-tanx dx\] equal to?

  39. anilorap
    • 4 years ago
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    ln/cosx/+c

  40. anilorap
    • 4 years ago
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    becuase the negative sign will change to postive once u take the derivative of cosx

  41. anilorap
    • 4 years ago
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    why u pic u =sinx?

  42. myininaya
    • 4 years ago
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    accident

  43. myininaya
    • 4 years ago
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    u=cosx du=-sinx dx

  44. anilorap
    • 4 years ago
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    :)

  45. anilorap
    • 4 years ago
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    good....:

  46. myininaya
    • 4 years ago
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    you want to try one i made up? its not hard

  47. anilorap
    • 4 years ago
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    ok

  48. myininaya
    • 4 years ago
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    \[\int\limits_{}^{}\frac{x+1}{x^2+2x}dx\]

  49. anilorap
    • 4 years ago
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    ok.. let me try

  50. anilorap
    • 4 years ago
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    u=x^2 + 2x du=2x+2 dx =2(x+1) dx so, 1/2du=x+1 dx

  51. myininaya
    • 4 years ago
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    very good so far

  52. anilorap
    • 4 years ago
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    1/2 integral of du/u =1/2 lnlul+c =1/2 ln/x^2+2xl+c

  53. myininaya
    • 4 years ago
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    :)

  54. anilorap
    • 4 years ago
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    yeaaaaaaaaaaaaa

  55. anilorap
    • 4 years ago
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    i love this stuff

  56. myininaya
    • 4 years ago
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    lol it is nice

  57. myininaya
    • 4 years ago
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    \[\int\limits_{}^{}\frac{f'(x)}{f(x)}dx\] so when we have this what will out answer be in the form of?

  58. anilorap
    • 4 years ago
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    f(x)*f'(x)

  59. myininaya
    • 4 years ago
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    look at the one you just did didn't we have f'/f dx?

  60. anilorap
    • 4 years ago
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    ohh... sorry.. yes yes

  61. myininaya
    • 4 years ago
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    =ln|f(x)|+C

  62. anilorap
    • 4 years ago
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    yes

  63. myininaya
    • 4 years ago
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    \[\int\limits_{}^{}[f(x)]^n*f'(x)dx\] \[n \neq -1\] what about this? what will our answer be?

  64. myininaya
    • 4 years ago
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    i will give you a hint let u=f(x) so du=f'(x) dx

  65. anilorap
    • 4 years ago
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    jjajjjaj... nice hint

  66. myininaya
    • 4 years ago
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    where are you from? you say jajaja alot lol

  67. anilorap
    • 4 years ago
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    dominican republic

  68. anilorap
    • 4 years ago
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    i am just laughing

  69. myininaya
    • 4 years ago
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    jajaaja= laughing?

  70. anilorap
    • 4 years ago
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    yess... how can i laugh??

  71. myininaya
    • 4 years ago
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    its fine the way you type your laughter no worries i like it

  72. anilorap
    • 4 years ago
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    lol

  73. myininaya
    • 4 years ago
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    so this all makes a little more sense?

  74. myininaya
    • 4 years ago
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    the integral stuff?

  75. anilorap
    • 4 years ago
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    why?

  76. myininaya
    • 4 years ago
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    im asking do you get the integral stuff better?

  77. anilorap
    • 4 years ago
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    ohhhhhh,, yess

  78. myininaya
    • 4 years ago
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    good :) im fixing to go to sleep goodnight

  79. anilorap
    • 4 years ago
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    for me is mornig.. but good nigth,,,

  80. myininaya
    • 4 years ago
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    good morning* i didn't sleep all night lol

  81. anilorap
    • 4 years ago
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    jajajajaaja

  82. anilorap
    • 4 years ago
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    ayayay.. no good

  83. anilorap
    • 4 years ago
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    thank u very much...

  84. myininaya
    • 4 years ago
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    np if you want to try another this is a good one \[\int\limits_{}^{}x*\sqrt{x+1} dx\] the answer is given above in one of the other threads if you want to check yourself

  85. myininaya
    • 4 years ago
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    later

  86. anilorap
    • 4 years ago
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    ok,, thanks

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