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

 

[Dick]

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

 

[Mark44]

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?

 

[Dick]

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]

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

 

[Dick]

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

Just factor (x^2+4)^(3/2) out of both terms in your solution to get a form like that.