# Indefinite Integrals & Substitution Rule

## Homework Statement

2. The attempt at a solution

I

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Dick
Homework Helper
It's not even close. Since when does the integral of du/sqrt(u) have an ln in it? And shouldn't you be using du=(-5)dx to replace dx with du?

It's not even close. Since when does the integral of du/sqrt(u) have an ln in it? And shouldn't you be using du=(-5)dx to replace dx with du?
I forgot about the square root.

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
Homework Helper
I forgot about the square root.

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)]
du/sqrt(u) is u^(-1/2)du. You've still got the power rule wrong.

Hurkyl
Staff Emeritus
Gold Member
Do remember there is a way to check your work -- what relationship is a function supposed to have with its anti-derivatives?

Do remember there is a way to check your work -- what relationship is a function supposed to have with its anti-derivatives?