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Math Help - find anti derivative of

  1. #1
    Member sluggerbroth's Avatar
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    find anti derivative of

    (1+x^2)/sq root x=.....?
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  2. #2
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    Re: find anti derivative of

    Quote Originally Posted by sluggerbroth View Post
    (1+x^2)/sq root x=.....?
    \displaystyle \begin{align*} \int{\frac{1 + x^2}{\sqrt{x}}\,dx} &= 2\int{\frac{1 + x^2}{2\sqrt{x}}\,dx} \end{align*}

    Let \displaystyle \begin{align*} u = \sqrt{x} \implies du = \frac{1}{2\sqrt{x}}\,dx \end{align*} and the integral becomes

    \displaystyle \begin{align*} 2\int{1 + u^4\,du} &= 2\left(u + \frac{u^5}{5}\right) + C \\ &= 2u + \frac{2u^5}{5} + C \\ &= 2\sqrt{x} + \frac{2\left(\sqrt{x}\right)^5}{5} + C \end{align*}
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  3. #3
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    Re: find anti derivative of

    Hello, sluggerbroth!

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

    We have: . \int\frac{1+x^2}{x^{\frac{1}{2}}}\,dx \;=\;\int\left(\frac{1}{x^{\frac{1}{2}}} + \frac{x^2}{x^{\frac{1}{2}}}\right)dx \;=\;\int\left(x^{-\frac{1}{2}} + x^{\frac{3}{2}}\right)dx

    Got it?
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