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anonymous

  • one year ago

Help finding the indefinite integral of one over sqrt of 1 + sqrt x

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  1. anonymous
    • one year ago
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    figure??

  2. anonymous
    • one year ago
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    \[\int\limits_{}^{} \frac{1}{\sqrt{1+\sqrt{x}}}\] And i know i'm gonna need trig functions.

  3. anonymous
    • one year ago
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    take root x = u^2

  4. anonymous
    • one year ago
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    \[x^{1/2} = u^{2} \]

  5. anonymous
    • one year ago
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    then differentiate \[x^{1/2}\] and get back to me

  6. anonymous
    • one year ago
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    |dw:1433523869741:dw|

  7. anonymous
    • one year ago
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    @cambrige okay I got the derivative of x^(1/2)|

  8. anonymous
    • one year ago
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    so dx= ur ans??

  9. anonymous
    • one year ago
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    My answer was 1 over 2root x. dx

  10. anonymous
    • one year ago
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    I'm lost now.

  11. anonymous
    • one year ago
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    @Concentrationalizing excuse me, if you could would you please help me with this problem?

  12. anonymous
    • one year ago
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    Consider u = 1 + \(\sqrt{x}\) du = \(\frac{1}{2\sqrt{x}}\)dx \(2\sqrt{x}du\) = dx Now, also notice from our substitution that \(\sqrt{x} = u-1\) Therefore we have \(2(u-1)du = dx\) Substituting these values into the integral gives: \[\int\limits_{}^{}\frac{ 1 }{ \sqrt{1+\sqrt{x}} }dx = \int\limits_{}^{}\frac{ 2(u-1) }{ \sqrt{u} }du\] Now we simply split this into two integrals and solve each. \[\int\limits_{}^{}\frac{ 2u-2 }{ \sqrt{u} }du = \int\limits_{}^{}\frac{ 2u }{ \sqrt{u} }du - \int\limits_{}^{}\frac{ 2 }{ \sqrt{u} }du\] And of course the left integral reduces and we have these two integrals \[2\int\limits_{}^{}\sqrt{u}du- 2\int\limits_{}^{}\frac{ 1 }{ \sqrt{u} }du\] Now just integrate each and plug back in your substitution from there. Think you can handle the integration from here? :) @study_guy2

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