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Integral calculation problem

  1. Jan 12, 2014 #1
  2. jcsd
  3. Jan 12, 2014 #2

    Dick

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    ##\frac{1}{3 ln(x+1/3)}## isn't right. ##\frac{1}{3} ln(x+1/3)## is right. Which one are your trying to write anyway? Use parentheses if you don't want to use TeX!
     
  4. Jan 12, 2014 #3
    Yes, the latter is the one! But, the computer program says it is wrong! Am I going crazy?
     
  5. Jan 12, 2014 #4

    Simon Bridge

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    the wolfram alpha input (link) was: integral of 1/(3x+1)=1/3ln(x+1/3)
    the output actually tells you why they think this relation is false.

    Namely: ##\ln\big(3x+1\big)\neq\ln\big(x+\frac{1}{3}\big)##

    But it doesn't have to be - equal, that is - since this is an indefinite integral, the two proposed solutions need only be the same to within an arbitrary constant. This is easy to show:

    ##\ln[x+\frac{1}{3}]=\ln[\frac{1}{3}(3x+1)]=\ln[3x+1]-\ln(3) = \ln[3x+1]+c##

    ... you have to be careful with indefinite integrals.
     
  6. Jan 12, 2014 #5
    Awesome! That's what's been missing! I spent an hour just on this!
     
  7. Jan 12, 2014 #6

    Simon Bridge

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    Don't trust the machines.

    You could have seen their result by using the substitution u=3x+1.
     
  8. Jan 12, 2014 #7
    Exactly, both are two acceptable results!
     
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