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Math Help - evaluate the integral

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    evaluate the integral

    integrate 1/(square root x + x^3/2)
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    Quote Originally Posted by twilightstr View Post
    integrate 1/(square root x + x^3/2)
    Is it \frac{1}{\sqrt{x} + x^{3/2}} ? If so, start by substituting u = \sqrt{x} \Rightarrow dx = 2 \sqrt{x} \, du.
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    no, entire the denominator is square root
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    Quote Originally Posted by twilightstr View Post
    no, entire the denominator is square root
    Then please make the effort to use appropriate grouping symbols.

    \int\frac{dx}{\sqrt{x + x^{3/2}}}

    =\int\frac{dx}{\sqrt{x(1 + x^{1/2})}}

    =\int\frac{dx}{\sqrt x\sqrt{1 + x^{1/2}}}

    =\int\frac{x^{-1/2}\,dx}{\sqrt{1 + x^{1/2}}}

    =2\int\frac{(1/2)x^{-1/2}\,dx}{\sqrt{1 + x^{1/2}}}

    You should be able to get it from here. Use u=1+x^{1/2}.
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