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Math Help - Definite integral ln(1+t)dt

  1. #1
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    Definite integral ln(1+t)dt

    I need some help with a particular question that i can't seem to figure out no matter how much time i've spent on it this evening..

    \int_{0}^{5} ln(1+t)dt

    My first way of thinking led me to a dead end. I used integration by parts with  U=ln(1+t) and  U`=1/(1+t) and V`=1dt and  V=t

    This led me to the following

     = [tln(1+t)]_{0}^{5} - \int_{0}^{5} t/(1+t)dt

    and this is where i'm stuck.. can't seem to figure out the integration for the second integral here.

    I think my line of thinking is wrong.. is my assumption that  U=ln(1+t) and  U`=1/(1+t) incorrect?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mkelly09 View Post
    I need some help with a particular question that i can't seem to figure out no matter how much time i've spent on it this evening..

    \int_{0}^{5} ln(1+t)dt

    My first way of thinking led me to a dead end. I used integration by parts with  U=ln(1+t) and  U`=1/(1+t) and V`=1dt and  V=t

    This led me to the following

     = [tln(1+t)]_{0}^{5} - \int_{0}^{5} t/(1+t)dt

    and this is where i'm stuck.. can't seem to figure out the integration for the second integral here.

    I think my line of thinking is wrong.. is my assumption that  U=ln(1+t) and  U`=1/(1+t) incorrect?
    You're correct...but you need to think a little "outside of the box" for this integral.

    \int\frac{t}{1+t}\,dt

    Let u=t+1\implies t=u-1

    Thus, \,du=\,dt

    Therefore, the integral becomes \int\frac{t\,du}{u}=\int\frac{u-1}{u}\,du=\int\left(1-\frac{1}{u}\right)\,du

    Can you take it from here?

    --Chris
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    You're correct...but you need to think a little "outside of the box" for this integral.

    \int\frac{t}{1+t}\,dt

    Let u=t+1\implies t=u-1

    Thus, \,du=\,dt

    Therefore, the integral becomes \int\frac{t\,du}{u}=\int\frac{u-1}{u}\,du=\int\left(1-\frac{1}{u}\right)\,du

    Can you take it from here?

    --Chris
    Here's another way I just thought of: add "zero" to the numerator

    \int\frac{t}{t+1}\,dt=\int\frac{t+1-1}{t+1}\,dt=\int\left[\frac{t+1}{t+1}-\frac{1}{t+1}\right]\,dt=\int\left(1-\frac{1}{t+1}\right)\,dt

    --Chris
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    using integration by parts on that guy directly is overkill. (that is, if you know the integral of ln(x) by heart--which you should).

    \int_0^5 \ln (1 + t)~dt

    Let u = t + 1, this substitution yields

    \int_1^6 \ln u~du

    which we know is u \ln u - u \bigg|_1^6 ...

    see post #7 here for how to integrate ln(x) using integration by parts. note that you could also figure out the integral by contemplating the derivative of xln(x)
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  5. #5
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    Thanks for your replies, they certainly solved my problem. In fact, now i know of a few different ways of doing it when before i couldn't figure out one.

    I think i will be committing to memory the antiderivative of ln(x)
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