1. ## Integration

∫x^2/√(x^2 + 1) dx

How can I do this?

2. Originally Posted by geton
∫x^2/√(x^2 + 1) dx

How can I do this?
Integration by parts........

3. Originally Posted by geton
∫x^2/√(x^2 + 1) dx

How can I do this?
$
\int {\frac{{x^2 }}
{{x^2 + 1}}\;dx = } \int {\frac{{x^2 + 1 - 1}}
{{x^2 + 1}}\;dx} = \int {1 - \frac{1}
{{x^2 + 1}}\;dx} \ldots$

edit: oops...missed your $\sqrt{}$...you need integration by parts as "mathstud" showed.

4. Yes I got it. Thanks.

5. Originally Posted by xifentoozlerix
$
\int {\frac{{x^2 }}
{{x^2 + 1}}\;dx = } \int {\frac{{x^2 + 1 - 1}}
{{x^2 + 1}}\;dx} = \int {1 - \frac{1}
{{x^2 + 1}}\;dx} \ldots$

edit: oops...missed your $\sqrt{}$...you need integration by parts as "mathstud" showed.
did you notice there is a square root in the denominator?

ok, i just saw your edit

6. alternatively, we could make a substitution, $u^2 = x^2 + 1$

or do a trig substitution, $x = \tan \theta$

7. Originally Posted by Jhevon
alternatively, we could make a substitution, $u^2 = x^2 + 1$
Trig. sub. will do it, but I'm afraid that your first sub. won't work.

Or make $\frac{{x^2 }}
{{\sqrt {x^2 + 1} }} = \sqrt {x^2 + 1} - \frac{1}
{{\sqrt {x^2 + 1} }}.$

You can tackle the first one via trig. - hyperbolic substitution or integration by parts, the another one is a classic one.

8. Originally Posted by Krizalid
Trig. sub. will do it, but I'm afraid that your first sub. won't work.

Or make $\frac{{x^2 }}
{{\sqrt {x^2 + 1} }} = \sqrt {x^2 + 1} - \frac{1}
{{\sqrt {x^2 + 1} }}.$

You can tackle the first one via trig. - hyperbolic substitution or integration by parts, the another one is a classic one.
hehe, i figure there might have been a problem with the first substitution, but i was too lazy to check. that's why i also put the trig sub there (i hoped to avoid trig sub altogether--you're rubbing off on me )