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Math Help - 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 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 u = \sqrt{x} \Rightarrow dx = 2u \, du.
    After which use some simple algebra to get 2 \int 1 - \frac{1}{u+1} \, du.
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  4. #4
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    <br />
\begin{gathered}<br />
  \int {\frac{4}<br />
{{4\sqrt x  + 4}}dx}  \hfill \\<br />
  \int {\frac{{dx}}<br />
{{\sqrt x  + 1}}}  \hfill \\<br />
  u = \sqrt x  \hfill \\<br />
  du = \tfrac{{dx}}<br />
{{2\sqrt x }} = \tfrac{{dx}}<br />
{{2u}} \Rightarrow dx = 2udu \hfill \\ <br />
\end{gathered} <br />

    <br />
2\int {\frac{{udu}}<br />
{{u + 1}}}  = 2\int {\frac{{u + 1 - 1}}<br />
{{u + 1}}du = 2\int {1 - \frac{1}<br />
{{u + 1}}du = 2u - 2\ln |u + 1| + C} } <br />

    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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