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

1. anonymous

figure??

2. anonymous

$\int\limits_{}^{} \frac{1}{\sqrt{1+\sqrt{x}}}$ And i know i'm gonna need trig functions.

3. anonymous

take root x = u^2

4. anonymous

$x^{1/2} = u^{2}$

5. anonymous

then differentiate $x^{1/2}$ and get back to me

6. anonymous

|dw:1433523869741:dw|

7. anonymous

@cambrige okay I got the derivative of x^(1/2)|

8. anonymous

so dx= ur ans??

9. anonymous

My answer was 1 over 2root x. dx

10. anonymous

I'm lost now.

11. anonymous

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