## anonymous 4 years ago antiderivative of x^2/ sqr(1+x^2)

1. Mr.Math

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

2. anonymous

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

3. anonymous

yes

4. Mr.Math

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, $$x=\tan(z)$$.

5. anonymous

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

6. Mr.Math

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

7. Mr.Math

Oh man!

8. anonymous

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

9. Mr.Math

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

10. anonymous

$\frac{x^2+ 1-1 }{\sqrt{1-x^2}}$ $-\sqrt{1-x^2} + \frac{1}{\sqrt{1-x^2}}$ $x = \sin\theta$ $dx = \cos\theta d\theta$ $-\cos^2\theta + 1$

11. anonymous

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