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Math Help - Need help with integration (questions with fractions, and e)

  1. #1
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    Need help with integration (questions with fractions, and e)

    Hi, I can do basic integration but on stuck on a few questions, with fractions and e. If you show me how to do these two, I'm sure I could do the rest.

    1. dx

    2. dx

    Thanks, really appreciate any input.
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  2. #2
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    Re: Need help with integration (questions with fractions, and e)

    Hello

    1. \mathrm{\displaystyle\int\dfrac{1}{4x+1}dx=\dfrac{  \ln(4x+1)}{4}+C}

    \boxed{\mathrm{\displaystyle\int\dfrac{dt}{t}=\ln t+C}}

    2. \mathrm{\displaystyle\int4e^{1-4x}=-e^{1-4x}+C}

    \boxed{\mathrm{\displaystyle\int e^t=e^t+C}}

    Greetings
    Last edited by darthjavier; November 13th 2012 at 03:46 AM.
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  3. #3
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    Re: Need help with integration (questions with fractions, and e)

    Quote Originally Posted by darthjavier View Post
    Hello

    1. \mathrm{\displaystyle\int\dfrac{1}{4x+1}dx=\dfrac{  \ln(4x+1)}{4}+C}

    \boxed{\mathrm{\displaystyle\int\dfrac{dt}{t}=\ln t+C}}

    2. \mathrm{\displaystyle\int4e^{1-4x}=-e^{1-4x}+C}

    \boxed{\mathrm{\displaystyle\int e^t=e^t+C}}

    Greetings
    No, the first is incorrect. It should be \displaystyle \begin{align*} \int{\frac{1}{4x + 1}\,dx} = \frac{\ln{|4x + 1|}}{4} + C \end{align*} since \displaystyle \begin{align*} \int{\frac{1}{t}\,dt} = \ln{|t|} + C \end{align*}.
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    Re: Need help with integration (questions with fractions, and e)

    Quote Originally Posted by Prove It View Post
    No, the first is incorrect. It should be \displaystyle \begin{align*} \int{\frac{1}{4x + 1}\,dx} = \frac{\ln{|4x + 1|}}{4} + C \end{align*} since \displaystyle \begin{align*} \int{\frac{1}{t}\,dt} = \ln{|t|} + C \end{align*}.
    Oops I'm sorry :P
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