- #1

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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

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- Thread starter charbon
- Start date

- #1

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

I have no idea how to substitute this integral in my favor. Can someone please set me on the right track? Thanks

- #2

Dick

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Substitute u^2=x^2+4. It's actually not so bad.

- #3

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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

Mark44

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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

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I'm sorry but I'm really clueless.

What step in my substitution did I do wrong?

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

Mark44

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Your work is correct - it's mine that is in error. Sorry for giving bad advice.I'm sorry but I'm really clueless.

What step in my substitution did I do wrong?

- #8

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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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