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    I'm aware this can be done by substitution or recognition. However, when I try and do this by parts, I get the integral in terms of itself (I = 1+ I +c) which then cancels out to give an error unless c=-1. Why can I not do this by parts?
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    (Original post by takeonme79)
    I'm aware this can be done by substitution or recognition. However, when I try and do this by parts, I get the integral in terms of itself (I = 1+ I +c) which then cancels out to give an error unless c=-1. Why can I not do this by parts?
    By recognition will be ln|lnx| because (ln(x))'=1/x
    By substitution: recognize what to substitute and what with
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    Yes, I know what the answer is by substitution or recognition but I would like to know why parts fails for this.
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    When you put in limits you'll find that the 1 part of that expression evaluates to zero.  1^{b}_{a} = 0

    Be aware that indefinite integrals take a little more thinking about when integrating by parts.

    So you end up with I = I
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    Thanks, I wasn't aware that when getting a constant like the one above in an indefinite integration by parts, that the constant evaluates to 0. What an evil trick!
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    (Original post by takeonme79)
    Thanks, I wasn't aware that when getting a constant like the one above in an indefinite integration by parts, that the constant evaluates to 0. What an evil trick!
    It's not about the constant evaluating to anything, it's just that if you put in limits then all of the constant terms automatically cancel. It's an arbitrary constant of integration -- it can take any value (the value may be specified, e.g. by saying that the graph of the integral passes through a certain point). So if you integrate something using two different methods and you end up with two functions that differ by a constant, then that's fine, because the integration constant is arbitrary and so when limits are applied you get the same answer.

    Take \sin x \cos x for example. Now, \dfrac{d}{dx} \sin^2 x = 2\sin x \cos x and \dfrac{d}{dx} \cos^2 x = -2\sin x \cos x, and so:

    \displaystyle \int \sin x \cos x\, dx = \dfrac{1}{2}\sin^2 x + C

    or alternatively

    \displaystyle \int \sin x \cos x\, dx = -\dfrac{1}{2}\cos^2 x + C

    The issue arises because the "+C" you add needn't be the same in each case. [Note that the two functions here do only differ by a constant because of the identity \cos^2 x = 1-\sin^2 x.]

    I've only gone into so much detail here because this issue comes up a lot and catches people off-guard.
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    Sorry, i didn't mean the constant of integration. i meant that if you attempt integration by parts on the integral of 1/(x ln x) you get the answer in the form I = 1 + I. I was referring to the constant 1.
 
 
 
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