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Math Help - Integration of Nat. Log of (x+1)

  1. #1
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    Integration of Nat. Log of (x+1)

    Hi. I would appreciate it if someone could show me how to integrate Ln (x+1). I have done integration by parts once arrived at xLn (x+1) minus the integration of x over (x+1).

    Many thanks
    Jay
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    MHF Contributor chisigma's Avatar
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    Is...

    \int \ln t \cdot dt = t\cdot \ln t - t + c (1)

    Now set in (1) 1+x instead of t ...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by jaijay32 View Post
    Hi. I would appreciate it if someone could show me how to integrate Ln (x+1). I have done integration by parts once arrived at xLn (x+1) minus the integration of x over (x+1).

    Many thanks
    Jay
    Split the fraction \frac x{x+1} = 1-\frac1{x+1}. The sum at the RHS can be integrated quite easily.
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  4. #4
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    \int \frac{x}{x+1} \, dx = \int \frac{x+1-1}{x+1} \, dx = \int \left( 1 - \frac{1}{x+1} \right) \, dx = \int \, dx - \int \frac{1}{x+1} \, dx
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    Integration by parts is correct.

    Quote Originally Posted by jaijay32 View Post
    Hi. I would appreciate it if someone could show me how to integrate Ln (x+1). I have done integration by parts once arrived at xLn (x+1) minus the integration of x over (x+1).

    Many thanks
    Jay
    I'll start from zero just so that is clearer:

     <br />
\int \ln{(x+1)} dx = (\int dx) \ln{(x+1)} - \int \frac{x}{x + 1} dx<br />

    Now I suppose this is what you have done so far.

    Try using the following equivalence ( which is obtained by dividing x by x+1):

     <br />
\frac{x}{x + 1} = 1 - \frac{1}{x + 1}<br />
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    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by jaijay32 View Post
    Hi. I would appreciate it if someone could show me how to integrate Ln (x+1). I have done integration by parts once arrived at xLn (x+1) minus the integration of x over (x+1).

    Many thanks
    Jay
    This is a familiar integral

    \int lnx dx= xlnx - x + c

    Does the addition of 1 change this?

    U=ln(x+1)

    dU = \frac{1}{x+1}

    dV=dx

    V=x

    \int ln(x+1) dx= xln(x+1) - \int \frac{x}{x+1}dx

    Which is where you got. Let us look at the integral then.

    \int \frac{x}{x+1}dx = \int \frac{x+1-1}{x+1}dx = \int 1 - \frac{1}{x+1} dx

    Of course this this the same as

    \int 1 - \frac{1}{x+1}dx = x-lnx + c

    We then get

    \int ln(x+1) dx= xlnx - x + lnx + c

    Edit- LOL at how many people jumped on answering this guy's question. I should quit here there's too many of us!
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    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by tom@ballooncalculus View Post
    That picture ****s me up hard. Can you describe that please lol
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  9. #9
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    ... is integration by parts, where



    ... is the product rule.

    Straight continuous lines differentiate downwards (integrate up) with respect to x. Choosing legs crossed or un-crossed does for assigning u du and v dv.

    Hope it helps, or amuses - but use whatever works for you.
    Cheers.
    _________________________________________
    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
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