Question

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

Answer 1

figure??

Answer 2

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

Answer 3

take root x = u^2

Answer 4

\[x^{1/2} = u^{2} \]

Answer 5

then differentiate \[x^{1/2}\]
and get back to me

Answer 6

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

Answer 7

so dx= ur ans??

Answer 8

My answer was 1 over 2root x. dx

Answer 9

I'm lost now.

Answer 10

@Concentrationalizing excuse me, if you could would you please help me with this problem?

Answer 11

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