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Math Help - integrate (ln(x+t)) dt

  1. #1
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    integrate (ln(x+t)) dt

    integrate (ln(x+t)) dt - Wolfram|Alpha



    how did it move from the penultimate step to the last?
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  2. #2
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    Re: integrate (ln(x+t)) dt

    Quote Originally Posted by BabyMilo View Post
    integrate (ln(x+t)) dt - Wolfram|Alpha



    how did it move from the penultimate step to the last?
    Before answering that, is x a function of t? Or is it being treated as constant?
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    Re: integrate (ln(x+t)) dt



    I am solving for this.
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    Re: integrate (ln(x+t)) dt

    Quote Originally Posted by Prove It View Post
    Before answering that, is x a function of t? Or is it being treated as constant?
    What difference does it make :S
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    Re: integrate (ln(x+t)) dt

    Quote Originally Posted by BabyMilo View Post
    integrate (ln(x+t)) dt - Wolfram|Alpha



    how did it move from the penultimate step to the last?
    Assuming that x is a constant value :

    first : substitute x+t=p , so dt=dp

    Hence :

    I=\int \ln p \,dp

    Now use integration by parts to solve this integral . Hint :

    u=\ln p ~\text{ and }~ dv=dp
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    Re: integrate (ln(x+t)) dt

    Quote Originally Posted by princeps View Post
    Assuming that x is a constant value :

    first : substitute x+t=p , so dt=dp

    Hence :

    I=\int \ln p \,dp

    Now use integration by parts to solve this integral . Hint :

    u=\ln p ~\text{ and }~ dv=dp
    you wrote what wolfram wrote in different symbols, didnt you?

    anyway, my question is how did it move from the penultimate step to the last?



    to



    Many thanks.
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    Re: integrate (ln(x+t)) dt

    Quote Originally Posted by BabyMilo View Post
    you wrote what wolfram wrote in different symbols, didnt you?

    anyway, my question is how did it move from the penultimate step to the last?



    to



    Many thanks.
    Actually solution is :

    I=(t+x)(\ln(t+x)-1)+C

    Hence :

    I=(t+x)\cdot \ln (t+x)-t+(C-x)

    Since x is a constant value we have that :

    I=(t+x)\cdot \ln (t+x)-t+C'
    Thanks from BabyMilo
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