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

  • Thread starter charbon
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  • #1
charbon
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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

  • #2
Dick
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Substitute u^2=x^2+4. It's actually not so bad.
 
  • #3
charbon
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Thanks for the help Dick.

I did what you suggested but I didn't quite get the correct answer.

Here are my steps:

[tex]\int[/tex]x^3[tex]\sqrt{(x^2+4)}[/tex]dx = [tex]\int[/tex](x^3(x^2+4))dx/[tex]\sqrt{(x^2+4)}[/tex]

u^2 = x^2+4 ; du = xdx/[tex]\sqrt{(x^2+4)}[/tex]

[tex]\int[/tex]x^2u^2du

x^2 = u^2 - 4

[tex]\int[/tex] (u^2-4)u^2du
= 1/5 (x^2+4)^5/2 - 4/3(x^2+4)^3/2 + c
 
  • #4
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Starting from [itex]\int x^3 (x^2 + 4)^{1/2}dx[/itex]

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 [itex]\int (u^2 - 4) u^2 du[/itex] shows that you have not substituted correctly. You should get two terms in u, both with fractional exponents.
 
  • #5
charbon
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I'm sorry but I'm really clueless.
What step in my substitution did I do wrong?
 
  • #6
Dick
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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?
 
  • #7
36,312
8,281
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.
 
  • #8
charbon
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Thank you both very much. The answer they give is [1/15(x^2+4)^3/2](3x^2-8) + c
 
  • #9
Dick
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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.
 

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