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Math Help - Integration of fractions to give ln quick help.

  1. #1
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    Integration of fractions to give ln quick help.

    Hi everyone - very quick one here

    Ill use f in bold do denote integration symbol because I havn't fully mastered the thing to enter formulae yet

    Anyway I'm looking to integrate [(-1)/(2-3x)]

    is it right to take 1/3 outside the integration symbol to give:

    +1/3f[(-1)/(2/3 -x)]
    (1/3) ln |(2/3)-x| + c
    or should it be:

    +1/3f[(-3)/(2-3x)
    (1/3) ln |2-3x| +c
    The answers gives the bottom one but I got the top one - am I wrong? why?

    Thanks
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Bladeowen View Post
    Hi everyone - very quick one here

    Ill use f in bold do denote integration symbol because I havn't fully mastered the thing to enter formulae yet

    Anyway I'm looking to integrate [(-1)/(2-3x)]

    is it right to take 1/3 outside the integration symbol to give:



    or should it be:


    The answers gives the bottom one but I got the top one - am I wrong? why?

    Thanks
    Are you trying to integrate [-1/(2-3x)] dx?

    If so, you CANNOT take 1/3 outside the integration symbol.

    Use Substution here:

    Suppose that u = 2-3x
    Then, du = -3 dx
    so, dx = -du/3

    substitute this value of dx in the question and you get,

    Integral of [-1/u] -du/3
    = Integration of 1/3 [1/u] du
    = 1/3* ln(u) +c
    substitute u = 2-3x, and you have
    1/3 * ln(2-3x) + c

    which is your answer...

    DO not forget to substitute in this case
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  3. #3
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    Ahh yes - many thanks
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  4. #4
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    Wait sorry can you please show me Integral of

    2/(1-x)

    In the same way? Sorry its basic stuff...
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Bladeowen View Post
    Wait sorry can you please show me Integral of

    2/(1-x)

    In the same way? Sorry its basic stuff...
    A lil bit of thinking and you are done!

    Suppose that u = 1-x
    Then, du = -dx
    so, dx = -du

    substitute this value of dx in the question

    You will be able to do the rest if you follow the previous answer!

    You should get:

    -2*ln(1-x)+C
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  6. #6
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    Oh sorry my mind was fried there - I'm an idiot

    I got dx=-du but forgot I was Integrating wrt u and not x so I put (1-x) back into where u was before I'd integrated!

    Idiot lol

    Thanks again
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