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## Homework Statement

**2. The attempt at a solution**

I

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- Thread starter MitsuShai
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I

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- #2

Dick

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I forgot about the square root.

how about this?

I= ∫ dx/sqrt(2−5x)

= ∫ [1/sqrt(2−5x)]dx

u= 2-5x

du= -5dx

(-1/5)du = dx

∫ (-1/5)du/sqrt(u)

(-1/5) ∫ du/u^(1/2)

(-1/5) 1/ (2/3)u^(3/2)

- 1/[(10/3)u^(3/2)]

- 1/[(10/3)(2-5x)^(3/2)]

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Dick

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du/sqrt(u) is u^(-1/2)du. You've still got the power rule wrong.I forgot about the square root.

how about this?

I= ∫ dx/sqrt(2−5x)

= ∫ [1/sqrt(2−5x)]dx

u= 2-5x

du= -5dx

(-1/5)du = dx

∫ (-1/5)du/sqrt(u)

(-1/5) ∫ du/u^(1/2)

(-1/5) 1/ (2/3)u^(3/2)

- 1/[(10/3)u^(3/2)]

- 1/[(10/3)(2-5x)^(3/2)]

- #5

Hurkyl

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- #6

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Yeah, I already know that.

I don't trust my math, that's why I want to know if I'm doing this correctly.

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- #7

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ok nevermind I got it.....

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