# Did I get the right anti derivate of this function?

1. Mar 1, 2012

So, I have a question about finding the integral of x^3(1+x^2)^1/2

This is basically what I did. Not sure if the answer is right or not.

I did the following u substitution.

u = 1 + x^2
u - 1 = x^2

du = 2xdx

x^2du/2 = 2xdu * x^2

x^2du/2 = x^3du

(u - 1)du/2*u^1/2

distributed the u

1/2*u^3/2 - u^1/2
u^2/2 = u

1/2 ∫ u = 1/4*u^2

= 1/4*(1+x^2)^2

2. Mar 1, 2012

### Staff: Mentor

No. You can always check that your antiderivative (there's no such term as antiderivate) is correct, by differenting it. When you do this, you should get the original integrand.

3. Mar 1, 2012

### Staff: Mentor

My first inclination, because of the square root of the sum of squares, would be a trig substitution. Another approach would be to use integration by parts.

If an ordinary substitution works, then that would be preferred, if you can find a suitable one, although I'm not sure that this the way to go in this problem.

Your work is hard to follow, so I didn't try to pick out where you went wrong. Make it easier to follow be starting with x3√(1 + x2) dx, and make you substitution in a systematic way.

4. Mar 1, 2012

I was sure I got this right. Well this is a bummer.

5. Mar 1, 2012

### Staff: Mentor

I would write this as
∫(1/2)(u - 1)u1/2du
= (1/2)∫(u3/2 - u1/2)du
Now, carry out the integration and undo your substitution.

6. Mar 2, 2012