- #1

Rijad Hadzic

- 321

- 20

## Homework Statement

Attached is a picture..

the book went from

[itex] (-1 + \frac {2}{1+u} ) du + \frac {dx}{x} = 0 [/itex]

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to

[itex] -u + 2ln(1+u) + ln(x) = ln(c) [/itex]

## Homework Equations

## The Attempt at a Solution

How in the world could the integral of 0 be ln(c)?

Maybe because it's 5 am and I'm up late and I'm hallucinating, but I always thought the integral of 0 is c, just a constant.

taking the integral of

[itex] (-1 \frac {2}{1+u} ) du + \frac {dx}{x} = 0 [/itex]

I get

[itex] (-u + 2 ln(u+1) + c ) + ln(x) + c_2 = c_3 [/itex]

[itex] -u +2ln(u+1) + ln(x) = c_4 [/itex]

#### Attachments

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