# Homework Help: Integral ∫x^3*√(x^2-5) dx

1. Nov 25, 2012

### ruiwp13

1. The problem statement, all variables and given/known data

∫x^3*√(x^2-5) dx

2. Relevant equations

∫u.dv=u.v-∫du.v

3. The attempt at a solution

So i tried to change the integral to ∫x*x^2*√(x^2-5)dx and u = x^2-5, then du = 2x, so 1/2*∫x^2*√(x^2-5) . Let u = √(x^2-5) , du = x/√(x^2-5) and dv = x^2 , v = x^3/3. Am I going in the right direction?

Last edited by a moderator: Feb 5, 2013
2. Nov 25, 2012

### Dick

Re: Integral

You don't need parts. Your first substitution is the one to use, u = x^2-5. Notice that means x^2=5+u.

3. Nov 25, 2012

### ruiwp13

Re: Integral

I did that, u=x^2-5, du= 2xdx, x^2=u+5 and I got :

1/2*∫(u+5)*√udu

so I multiplied and I got : 1/2*∫u^(3/2)+5u^(1/2)

1/2*1/5 * ∫u^(3/2)+u^(1/2)

and I got u^(5/2)/25+(2u^(3/2))/3 and It's wrong :/ where did I failed?

Thanks

4. Nov 25, 2012

### Dick

Re: Integral

Just sloppy algebra, I think. Where did the '1/5' come from?

5. Nov 25, 2012

### ruiwp13

Re: Integral

To remove the 5 from 5u^(1/2)

The solution is supposed to be:

x^3/3*(u^(3/2))-2/15*(u^(5/2))+c

6. Nov 25, 2012

### Dick

Re: Integral

'Remove the 5'?? I don't get it. And the given solution looks wrong as well.

7. Nov 25, 2012

### ruiwp13

8. Nov 25, 2012

### Dick

9. Nov 25, 2012

### ruiwp13

Re: Integral

So where is my mistake?

10. Nov 25, 2012

### Dick

Re: Integral

Start with 'removing the 5'. You can write u^(3/2)+5u^(1/2)=5*((1/5)*u^(3/2)+u^(1/2)). That's not (1/5)*(u^(3/2)+u^(1/2)), if that's what you meant. It's hard to tell how you are grouping things. Use more parentheses to make things clearer.