Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

ggrreeBest ResponseYou've already chosen the best response.0
\[\int\limits_{} {dx \over 1+ \sqrt x}\]
 2 years ago

myininayaBest ResponseYou've already chosen the best response.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...
 2 years ago

myininayaBest ResponseYou've already chosen the best response.2
We could try another way if you don't like this one...
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.1
\[\int\limits\frac{1}{1+\sqrt{x}} dx\] \[x = u^2 =>u=\sqrt{x} \] \[\frac{dx}{du} =2u =>dx=2u*du\] \[=>\int\limits\frac{1}{1+u} *2u*du =\int\limits\frac{u}{u+1} \] Might be easier..
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.1
woops, i made a typo, the last one should be.. \[2\int\limits\frac{u}{u+1} du\]
 2 years ago

ChlorophyllBest ResponseYou've already chosen the best response.0
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
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.