Question

antiderivative of
x^2/
sqr(1+x^2)

Answer 1

Is it \(\frac{x^2}{\sqrt{1+x^2}}?\)

Answer 2

\( \int \frac{x^2 dx}{\sqrt{1+x^2}} \)

Answer 3

yes

Answer 4

You know that the most general anti-derivative 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, \(x=\tan(z)\).

Answer 5

oh, I'm so sorry, insted of 1+x^2 is 1-x^2

Answer 6

\(x=\tan(z) \implies dx=\sec^2(z)dz\), the integral becomes:
\[\int\limits {\tan^2(z)\sec^2(z) dz \over \sec(z)}=\int\limits \tan^2(z)\sec(z)dz\]

Answer 7

Oh man!

Answer 8

if he at least gives me any idea to solve is already worth it :D

Answer 9

You will use the same method, but substitute \(x=\sin(z)\) instead!

Answer 10

\[\frac{x^2+ 1-1 }{\sqrt{1-x^2}}\]
\[-\sqrt{1-x^2} + \frac{1}{\sqrt{1-x^2}}\]

\[x = \sin\theta \]
\[dx = \cos\theta d\theta\]

\[-\cos^2\theta + 1\]

Answer 11

that'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