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    hi, The integral of 1/3x by the rule (1/a)ln(ax + b) would be 1/3ln3x. But if you take out a third at the beginning you would be left with 1/3 times integral of 1/x which is 1/3 lnx.

    What am i doing wrong?
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    (Original post by 111davey1)
    hi, The integral of 1/3x by the rule (1/a)ln(ax + b) would be 1/3ln3x. But if you take out a third at the beginning you would be left with 1/3 times integral of 1/x which is 1/3 lnx.

    What am i doing wrong?
    It's fine to do it either way. Just remember that indefinite integrals have a +C. In the two forms you've mentioned, the C values will be different as \dfrac{1}{3}\ln3x=\dfrac{1}{3} \ln3x+\underbrace{\frac{1}{3}\ln  3}_{\text{constant}}
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    (Original post by 111davey1)
    hi, The integral of 1/3x by the rule (1/a)ln(ax + b) would be 1/3ln3x. But if you take out a third at the beginning you would be left with 1/3 times integral of 1/x which is 1/3 lnx.

    What am i doing wrong?
    I guess it's safe to assume you mean \displaystyle \int \frac{1}{3x} .dx in which case, what rule do you mean?? Simply saying "\frac{1}{a}\ln(ax+b)" is not stating any 'rule'

    The integral of \displaystyle \int \frac{1}{3x} .dx = \frac{1}{3}\int \frac{1}{x} .dx = \frac{1}{3}\ln \vert x \vert +c

    If you have \displaystyle \int \frac{1}{ax+b} .dx then this is the same as \displaystyle \frac{1}{a} \int \frac{a}{ax+b} .dx = \frac{1}{a} \ln \vert ax+b \vert + c which you can surely apply by saying a=3 and b=0 which indeed gets you \frac{1}{3}\ln \vert 3x \vert+c but remember that you can split this into \frac{1}{3}[\ln(3)+\ln \vert x \vert] + c which then goes to:

    \frac{1}{3}\ln \vert x \vert + \frac{1}{3}\ln(3) + c = \frac{1}{3}\ln \vert x \vert + D for some arbitrary constant D - which is the same as the first approach.

    EDIT: Well I guess I've been beaten to it
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    (Original post by RDKGames)
    I guess it's safe to assume you mean \displaystyle \int \frac{1}{3x} .dx in which case, what rule do you mean?? Simply saying "\frac{1}{a}\ln(ax+b)" is not stating any 'rule'

    The integral of \displaystyle \int \frac{1}{3x} .dx = \frac{1}{3}\int \frac{1}{x} .dx = \frac{1}{3}\ln \vert x \vert +c

    If you have \displaystyle \int \frac{1}{ax+b} .dx then this is the same as \displaystyle \frac{1}{a} \int \frac{a}{ax+b} .dx = \frac{1}{a} \ln \vert ax+b \vert + c which you can surely apply by saying a=3 and b=0 which indeed gets you \frac{1}{3}\ln \vert 3x \vert+c but remember that you can split this into \frac{1}{3}[\ln(3)+\ln \vert x \vert] + c which then goes to:

    \frac{1}{3}\ln \vert x \vert + \frac{1}{3}\ln(3) + c = \frac{1}{3}\ln \vert x \vert + D for some arbitrary constant D - which is the same as the first approach.

    EDIT: Well I guess I've been beaten to it
    Thanks very much. Sorry i should have explained the 'rule'.
 
 
 

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