- #1

- 13

- 0

**[SOLVED] Integration, u substitution, 1/u**

-- +C at the end of the integral solutions, I can't seem to add it in the LaTeX thing --

## Homework Statement

#1 [tex]\int\frac{1}{8-4x}dx[/tex]

#2 [tex]\int\frac{1}{2x}dx[/tex]

## The Attempt at a Solution

#1

Rewrite algebraically:

[tex]\int\frac{1}{x-2}*\frac{-1}{4}dx[/tex]

Pull out constant:

[tex]\frac{-1}{4}\int\frac{1}{x-2}dx[/tex]

Usub, u=x-2, du=1*dx,

[tex]\frac{-1}{4}\int\frac{1}{u}du[/tex]

[tex]\frac{-ln(u)}{4}[/tex] -> Answer: [tex]\frac{-ln(x-2)}{4}[/tex]

#2

Pull out constant: [tex]\frac{1}{2}*\int\frac{1}{x}dx[/tex]

Answer: [tex]\frac{ln(x)}{2}[/tex]

--OR, if I don't pull out constants--

#1

Usub, u=8-4x, du=-4dx,

[tex]\frac{-1}{4}*\int\frac{1}{u}du[/tex]

[tex]\frac{-ln(u)}{4}[/tex] -> Answer: [tex]\frac{-ln(8-4x)}{4}[/tex], which =/= [tex]\frac{-ln(x-2)}{4}[/tex]

#2

Usub, u=2x, du = 2dx,

[tex]\frac{1}{2}[/tex][tex]\int\frac{1}{u}du[/tex]

[tex]\frac{ln(u)}{2}[/tex] -> Answer: [tex]\frac{ln(2x)}{2}[/tex], which =/= [tex]\frac{ln(x)}{2}[/tex]

I know the answers are -1/4*ln(8-4x) and ln(x)/2 respectively, but I don't understand why I cannot pull out the constant for the first one, yet I am required to pull it out for the second?

Bonus question: Why does my TI-89 integrate #1 incorrectly, but #2 correctly? :(

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