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    How would you intergrate (x+2)/(2x-3)^3 dx
    If u=2x-3
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    (Original post by toonervoustotalk)
    How would you intergrate (x+2)/(2x+3)^3 dx
    If u=2x-3
    Strange substitution choice. Is it not u = 2x+3?

    Anyway, what have you tried so far?
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    (Original post by SeanFM)
    Strange substitution choice. Is it not u = 2x+3?

    Anyway, what have you tried so far?
    Because i made a mistake in the denominator
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    (Original post by toonervoustotalk)
    Because i made a mistake in the denominator
    What is the mistake?

    And what working have you got so far for the question?
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    (Original post by toonervoustotalk)
    How would you intergrate (x+2)/(2x+3)^3 dx
    If u=2x-3
    I think you meant u=2x+3.
    Differentiate 2x+3 with respect to x and replace dx with 0.5du

    then replace x+2 with some multiple of u (hint)

    You should be able to integrate from there...
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    (Original post by toonervoustotalk)
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    You've replaced dx by du but forgot the /2.

    The 1/2 from (u+3)/2 has also disappeared in your substitution.
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    (Original post by SeanFM)
    You've replaced dx by du but forgot the /2.

    The 1/2 from (u+3)/2 has also disappeared in your substitution.
    I figured out after i sent it
    Name:  DSC_0351.jpg
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    This any better?
    I know i somehow have to get ln something as 1/x intergrates to lnx
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    (Original post by toonervoustotalk)
    I figured out after i sent it
    Name:  DSC_0351.jpg
Views: 79
Size:  474.9 KB
    This any better?
    I know i somehow have to get ln something as 1/x intergrates to lnx
    don't forget the du on the RHS. Re-arrange the fraction and integrate. You can take constants outside the integral if you want.
 
 
 
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Updated: October 20, 2015
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