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Math Help - Integration

  1. #1
    Member srirahulan's Avatar
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    Post Integration

    \int[\frac{ln(x+1)}{x+1}] How can i solve this for simple way. i think a solution for a longtime but i cannot get anything.please give me solution?????
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  2. #2
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    Re: Integration

    You forget dx in the integral (very important. You need it to the change of variable)
    Let y=ln(x+1)
    dy=dx/(x+1)
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  3. #3
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    Re: Integration

    Quote Originally Posted by srirahulan View Post
    \int[\frac{ln(x+1)}{x+1}] How can i solve this for simple way. i think a solution for a longtime but i cannot get anything.please give me solution?????
    What is the derivative of \tfrac{1}{2}\ln^2(x+1)~?
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    Re: Integration

    Integration-02-aug-2-13.png
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  5. #5
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    Re: Integration

    Quote Originally Posted by srirahulan View Post
    \int[\frac{ln(x+1)}{x+1}] How can i solve this for simple way. i think a solution for a longtime but i cannot get anything.please give me solution?????
    No, you won't be "given a solution", but you can be given some hints. First of all, having "x + 1" in both the numerator and denominator makes things look convoluted. So a substitution \displaystyle \begin{align*} u = x + 1 \implies du = dx \end{align*} would clean things up a bit, giving

    \displaystyle \begin{align*} \int{\frac{\ln{(x + 1)}}{x + 1}\,dx} &= \int{\frac{\ln{(u)}}{u}\,du} \\ &= \int{\ln{(u)}\cdot \frac{1}{u}\,du} \end{align*}

    Now notice that \displaystyle \begin{align*} \frac{1}{u} \end{align*} is the derivative of \displaystyle \begin{align*} \ln{(u)} \end{align*}, so a substitution of the form \displaystyle \begin{align*} v = \ln{(u)} \implies dv = \frac{1}{u}\,du \end{align*} is appropriate, giving

    \displaystyle \begin{align*} \int{\ln{(u)}\cdot \frac{1}{u}\,du} = \int{ v\,dv} \end{align*}

    Surely you can integrate THAT.
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