Question

integral of dx/[1+sqrt x]
(rewritten in reply)

Answer 1

\[\int\limits_{} {dx \over 1+ \sqrt x}\]

Answer 2

\[\int\limits_{}^{}\frac{1}{1+\sqrt{x}} dx\]
\[\text{ let } \sqrt{x}=\tan^2(\theta) \]
\[ x=\tan^4(\theta)\]
=>
\[dx=4 \tan^3(\theta) \sec^2(\theta) d \theta\]
So we have
\[\int\limits_{}^{}\frac{4 \tan^3(\theta) \sec^2(\theta)}{1+\tan^2(\theta)} d \theta\]
\[4 \int\limits_{}^{}\tan^3(\theta) d \theta\]
That should help...

Answer 3

We could try another way if you don't like this one...

Answer 4

\[\int\limits\frac{1}{1+\sqrt{x}} dx\]
\[x = u^2 =>u=\sqrt{x} \]
\[\frac{dx}{du} =2u =>dx=2u*du\]
\[=>\int\limits\frac{1}{1+u} *2u*du =\int\limits\frac{u}{u+1} \]

Might be easier..

Answer 5

yep that's the one

Answer 6

woops, sorry

Answer 7

woops, i made a typo, the last one should be..
\[2\int\limits\frac{u}{u+1} du\]

Answer 8

Let u = sqrtx -> u^2 = x => 2udu = dx
-> 2 Int ( udu/ u + 1)
-> 2 Int [ 1 - 1/(u + 1 )] du
= 2 [ u - ln(u + 1) ]
= 2 ( sqrtx - ln(sqrtx + 1 ) + C

Answer 9

thanks guys!