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Integral ∫x^3*√(x^2-5) dx

by ruiwp13
Tags: ∫x3√x25, integral
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ruiwp13
#1
Nov25-12, 12:17 PM
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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Dick
#2
Nov25-12, 12:36 PM
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Quote Quote by ruiwp13 View Post
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.
ruiwp13
#3
Nov25-12, 12:49 PM
P: 40
Quote Quote by Dick View Post
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

Dick
#4
Nov25-12, 12:56 PM
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Integral ∫x^3*√(x^2-5) dx

Quote Quote by ruiwp13 View Post
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?
ruiwp13
#5
Nov25-12, 12:59 PM
P: 40
Quote Quote by Dick View Post
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
Dick
#6
Nov25-12, 01:08 PM
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Quote Quote by ruiwp13 View Post
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.
ruiwp13
#7
Nov25-12, 01:11 PM
P: 40
Quote Quote by Dick View Post
'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
Dick
#8
Nov25-12, 01:21 PM
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Quote Quote by ruiwp13 View Post
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).
ruiwp13
#9
Nov25-12, 01:24 PM
P: 40
Quote Quote by Dick View Post
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?
Dick
#10
Nov25-12, 01:31 PM
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Quote Quote by ruiwp13 View Post
So where is my mistake?
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.


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