# Integral ∫x^3*√(x^2-5) dx

by ruiwp13
Tags: ∫x3√x25, integral
 P: 40 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? Thanks in advance
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P: 25,235
 Quote by 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? Thanks in advance
You don't need parts. Your first substitution is the one to use, u = x^2-5. Notice that means x^2=5+u.
P: 40
 Quote by Dick You don't need parts. Your first substitution is the one to use, u = x^2-5. Notice that means x^2=5+u.
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

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P: 25,235
Integral ∫x^3*√(x^2-5) dx

 Quote by ruiwp13 so I multiplied and I got : 1/2*∫u^(3/2)+5u^(1/2) 1/2*1/5 * ∫u^(3/2)+u^(1/2)
Just sloppy algebra, I think. Where did the '1/5' come from?
P: 40
 Quote by Dick Just sloppy algebra, I think. Where did the '1/5' come from?
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
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P: 25,235
 Quote by ruiwp13 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
'Remove the 5'?? I don't get it. And the given solution looks wrong as well.
P: 40
 Quote by Dick 'Remove the 5'?? I don't get it. And the given solution looks wrong as well.
Trough wolframalpha the solution is different:

http://www.wolframalpha.com/input/?i...t%28x%5E2-5%29
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P: 25,235
 Quote by ruiwp13 Trough wolframalpha the solution is different: http://www.wolframalpha.com/input/?i...t%28x%5E2-5%29
Wolfram is using u^(5/2)=u*u^(3/2) and pulling out the common factor of u^(3/2).
P: 40
 Quote by Dick Wolfram is using u^(5/2)=u*u^(3/2) and pulling out the common factor of u^(3/2).
So where is my mistake?