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anonymous
 5 years ago
antiderivative of
x^2/
sqr(1+x^2)
anonymous
 5 years ago
antiderivative of x^2/ sqr(1+x^2)

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

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

Mr.Math
 5 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
 5 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
 5 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
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.2You will use the same method, but substitute \(x=\sin(z)\) instead!

anonymous
 5 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
 5 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
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