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    Ok i keep using parts and this is pants

    im trying to integrate 1/(x+1) how can i do this???

    The Q. was dy/dx - y/(x+1) = x

    i got it down to

    d/dx(y(x+1)^-1) = x/(x+1) now im trying to integrate the RHS
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    It's just ln(x+1) + C.

    d[ln(ax+b)]/dx = a/(ax+b)
    => int 1/(ax+b) dx = (1/a) ln(ax+b) + C

    And: "pants"? :confused:
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    (Original post by dvs)
    It's just ln(x+1) + C.

    d[ln(ax+b)]/dx = a/(ax+b)
    => int 1/(ax+b) dx = (1/a) ln(ax+b) + C

    And: "pants"? :confused:









    oooops i meant x/(x+1) sowwy
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    x/(x+1) = (x+1-1)/(x+1) = (x+1)/(x+1) - 1/(x+1) = 1 - 1/(x+1)
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    (Original post by dvs)
    x/(x+1) = (x+1-1)/(x+1) = (x+1)/(x+1) - 1/(x+1) = 1 - 1/(x+1)
    therefore the integral is

    x-ln(x+1) ja?
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    (Original post by futureaussiecto)
    therefore the integral is

    x-ln(x+1) ja?
    Yes
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    Why would you need parts for this?

    As far as I'm concerned you only need parts if youre integrating a product of a polynomial AND ONE OF a trig, exponential or ln function.
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    (Original post by distortedgav)
    Why would you need parts for this?

    As far as I'm concerned you only need parts if youre integrating a product of a polynomial AND ONE OF a trig, exponential or ln function.
    It has other applications, e.g. for finding the integral of (e^x)(sin(x)) or (e^x)(cos(x)) as well as finding reduction formulae.
 
 
 
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