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Math Help - Log Integration

  1. #1
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    Log Integration

    Hi, this is a solution to a question that I couldn't do although I understand the rest I just don't get the following line \int_{0}^{1}\frac{1}{(x+4)}-\frac{4}{(x+4)^2}dx={[log (x+4)+\frac{4}{x+4}]}^{1}_{0} I get how the the \frac{1}{x+4} was integrated but I dont get the fraction next to it Thanks
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  2. #2
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    Because \displaystyle \int{\frac{4}{(x+4)^2}\,dx} = \int{4(x+4)^{-2}\,dx}

    \displaystyle = \int{4u^{-2}\,du} after making the substitution \displaystyle u = x+4

    \displaystyle = \frac{4u^{-1}}{-1} + C

    \displaystyle = -\frac{4}{(x+4)} + C.
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  3. #3
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    Quote Originally Posted by aonin View Post
    Hi, this is a solution to a question that I couldn't do although I understand the rest I just don't get the following line \int_{0}^{1}\frac{1}{(x+4)}-\frac{4}{(x+4)^2}dx={[log (x+4)+\frac{4}{x+4}]}^{1}_{0} I get how the the \frac{1}{x+4} was integrated but I dont get the fraction next to it Thanks
    \displaystyle \int -\frac{4}{(x+4)^2} \, dx = -4\int (x+4)^{-2} \, dx = 4(x+4)^{-1} = \frac{4}{x+4}

    in future, post calculus questions in the calculus forum.
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