# Integrate x^3sqrt(x^2+4)

## Homework Statement

Integrate x^3sqrt(x^2+4)

The attempt at a solution

I have no idea how to substitute this integral in my favor. Can someone please set me on the right track? Thanks

Dick
Homework Helper
Substitute u^2=x^2+4. It's actually not so bad.

Thanks for the help Dick.

I did what you suggested but I didn't quite get the correct answer.

Here are my steps:

$$\int$$x^3$$\sqrt{(x^2+4)}$$dx = $$\int$$(x^3(x^2+4))dx/$$\sqrt{(x^2+4)}$$

u^2 = x^2+4 ; du = xdx/$$\sqrt{(x^2+4)}$$

$$\int$$x^2u^2du

x^2 = u^2 - 4

$$\int$$ (u^2-4)u^2du
= 1/5 (x^2+4)^5/2 - 4/3(x^2+4)^3/2 + c

Mark44
Mentor
Starting from $\int x^3 (x^2 + 4)^{1/2}dx$

u^2 = x^2 + 4 ==> x^2 = u^2 - 4 ==> x = (u^2 - 4)^(1/2)
and udu = xdx

Your integral $\int (u^2 - 4) u^2 du$ shows that you have not substituted correctly. You should get two terms in u, both with fractional exponents.

I'm sorry but I'm really clueless.
What step in my substitution did I do wrong?

Dick
Homework Helper
I'm sorry but I'm really clueless.
What step in my substitution did I do wrong?

You didn't do anything wrong. Your answer is right. But there are other ways to write the answer. What answer are you expecting?

Mark44
Mentor
I'm sorry but I'm really clueless.
What step in my substitution did I do wrong?

Thank you both very much. The answer they give is [1/15(x^2+4)^3/2](3x^2-8) + c

Dick