A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
antiderivative of
x^2/
sqr(1+x^2)
anonymous
 4 years ago
antiderivative of x^2/ sqr(1+x^2)

This Question is Closed

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.2Is it \(\frac{x^2}{\sqrt{1+x^2}}?\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\( \int \frac{x^2 dx}{\sqrt{1+x^2}} \)

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.2You know that the most general antiderivative of it is: \[\int\limits {x^2 \over \sqrt{1+x^2}}dx\] The best way to evaluate this integral is by using trig substitution, \(x=\tan(z)\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh, I'm so sorry, insted of 1+x^2 is 1x^2

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.2\(x=\tan(z) \implies dx=\sec^2(z)dz\), the integral becomes: \[\int\limits {\tan^2(z)\sec^2(z) dz \over \sec(z)}=\int\limits \tan^2(z)\sec(z)dz\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if he at least gives me any idea to solve is already worth it :D

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.2You will use the same method, but substitute \(x=\sin(z)\) instead!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{x^2+ 11 }{\sqrt{1x^2}}\] \[\sqrt{1x^2} + \frac{1}{\sqrt{1x^2}}\] \[x = \sin\theta \] \[dx = \cos\theta d\theta\] \[\cos^2\theta + 1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's what i did but i'm kind of dyslexic so i must have mixed something :D i'll check again to see what's wrong
Ask your own question
Sign UpFind more explanations on OpenStudy
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.