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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{} \frac{1}{\sqrt{1+\sqrt{x}}}\] And i know i'm gonna need trig functions.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x^{1/2} = u^{2} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then differentiate \[x^{1/2}\] and get back to me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433523869741:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@cambrige okay I got the derivative of x^(1/2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My answer was 1 over 2root x. dx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Concentrationalizing excuse me, if you could would you please help me with this problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Consider u = 1 + \(\sqrt{x}\) du = \(\frac{1}{2\sqrt{x}}\)dx \(2\sqrt{x}du\) = dx Now, also notice from our substitution that \(\sqrt{x} = u1\) Therefore we have \(2(u1)du = dx\) Substituting these values into the integral gives: \[\int\limits_{}^{}\frac{ 1 }{ \sqrt{1+\sqrt{x}} }dx = \int\limits_{}^{}\frac{ 2(u1) }{ \sqrt{u} }du\] Now we simply split this into two integrals and solve each. \[\int\limits_{}^{}\frac{ 2u2 }{ \sqrt{u} }du = \int\limits_{}^{}\frac{ 2u }{ \sqrt{u} }du  \int\limits_{}^{}\frac{ 2 }{ \sqrt{u} }du\] 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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