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Thread: hard integral(i think)

  1. #1
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    hard integral(i think)

    integrate (4) / (4(x^1/2) + 4)

    please help me i cannot seem to get this
    i believe i need a u substitution, but i am not sure
    thanks in advance
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  2. #2
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    Quote Originally Posted by whocares27 View Post
    integrate (4) / (4(x^1/2) + 4)

    please help me i cannot seem to get this
    i believe i need a u substitution, but i am not sure
    thanks in advance
    Make the substitution $\displaystyle u = \sqrt{x} \Rightarrow dx = 2u \, du$.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Make the substitution $\displaystyle u = \sqrt{x} \Rightarrow dx = 2u \, du$.
    After which use some simple algebra to get $\displaystyle 2 \int 1 - \frac{1}{u+1} \, du$.
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  4. #4
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    $\displaystyle
    \begin{gathered}
    \int {\frac{4}
    {{4\sqrt x + 4}}dx} \hfill \\
    \int {\frac{{dx}}
    {{\sqrt x + 1}}} \hfill \\
    u = \sqrt x \hfill \\
    du = \tfrac{{dx}}
    {{2\sqrt x }} = \tfrac{{dx}}
    {{2u}} \Rightarrow dx = 2udu \hfill \\
    \end{gathered}
    $

    $\displaystyle
    2\int {\frac{{udu}}
    {{u + 1}}} = 2\int {\frac{{u + 1 - 1}}
    {{u + 1}}du = 2\int {1 - \frac{1}
    {{u + 1}}du = 2u - 2\ln |u + 1| + C} }
    $

    Then substitute the square root of x back in for u and your set.
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  5. #5
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    thank you both so very much
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