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

charbon

## 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

## Answers and Replies

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

charbon
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

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

make all your substitutions, using
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.

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

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?

Mentor
I'm sorry but I'm really clueless.
What step in my substitution did I do wrong?
Your work is correct - it's mine that is in error. Sorry for giving bad advice.

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