## ggrree Group Title integral of dx/[1+sqrt x] (rewritten in reply) 2 years ago 2 years ago

1. ggrree Group Title

$\int\limits_{} {dx \over 1+ \sqrt x}$

2. myininaya Group Title

$\int\limits_{}^{}\frac{1}{1+\sqrt{x}} dx$ $\text{ let } \sqrt{x}=\tan^2(\theta)$ $x=\tan^4(\theta)$ => $dx=4 \tan^3(\theta) \sec^2(\theta) d \theta$ 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...

3. myininaya Group Title

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

4. Mimi_x3 Group Title

$\int\limits\frac{1}{1+\sqrt{x}} dx$ $x = u^2 =>u=\sqrt{x}$ $\frac{dx}{du} =2u =>dx=2u*du$ $=>\int\limits\frac{1}{1+u} *2u*du =\int\limits\frac{u}{u+1}$ Might be easier..

5. myininaya Group Title

yep that's the one

6. Mimi_x3 Group Title

woops, sorry

7. Mimi_x3 Group Title

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

8. Chlorophyll Group Title

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

9. ggrree Group Title

thanks guys!