Question

How do I find the integral of 1/(sqrt(1+sqrt(x))) dx

Answer 1

Have you tried \(\large 1+\sqrt{x} = u^2\) ?

Answer 2

Ah I didnt think about to square u, let me try.

Answer 3

Uuhm, so \[du ^{2}=\frac{ dx }{ 2\sqrt{x} }\]

Answer 4

We didnt learn about squaring u's before so I'm not sure how it works excactly

Answer 5

no its 2du=dx/2x^1/2

Answer 6

Ok so I got \[2du=\frac{ dx }{ 2\sqrt{x} }\]

How do I get \[\sqrt{1+\sqrt{x}}\]

into \[\frac{ dx }{ 2\sqrt{x} }\]

Answer 7

I mean \[\frac{ dx }{ \sqrt{1+\sqrt{x}} }\]

Answer 8

\[u got the answer\]

Answer 9

I don't see how im there yet :P I need to get \[\frac{ dx }{ \sqrt{1+\sqrt{x}} }\] into \[\frac{ dx }{ 2\sqrt{x} }\]

Answer 10

\[\large 1+\sqrt{x} = u^2 \implies \dfrac{1}{2\sqrt{x}}dx = 2u du \implies dx = 4u(u^2-1)du\]

Answer 11

the integral becomes :

\[\large \int \dfrac{4u(u^2-1)du}{\sqrt{u^2}}\]

Answer 12

How did you got from \[\frac{ 1 }{ 2\sqrt{x} }dx =2udu\]

to

\[dx =4u(u^2-1)du\]

Answer 13

substitute \(\large \sqrt{x} = u^2-1\)

Answer 14

Ohyea ofcourse :P Thanks alot!!

Answer 15

Sorry but this one is really hard. Further I got the integral and is \[\frac{ 4 }{ 3 }u ^{3}-4x\] which becomes \[\frac{ 4 }{ 3 }(1+\sqrt{x})^{\frac{ 3 }{ 2 }}-4x\] Because \[u ^{2}=1+\sqrt{x}\] so \[u ^{3}=(1+\sqrt{x})^{\frac{ 3 }{ 2 }}\]

I must have done something wrongg but i dont knwo what