Different solutions to x/(1-x) by substitution and decomposition

[greentea]

Homework Statement

Derive x/(1-x)

Homework Equations

By substitution:
u = (1-x)
du = -dx
∫x/(1-x)dx = -∫(1-u)/u du = ∫1-(1/u)du = u-log(u)+c = (1-x) - log(1-x) + c

By decomposition:
x/(1-x) = 1/(1-x)-1
∫1/(1-x)-1dx = -log(1-x)-x+c

The Attempt at a Solution

Which solutions should I use (1-x) - log(1-x) + c or -log(1-x)-x+c, or are both equally good?

 

[Dick]

Homework Statement

Derive x/(1-x)

Homework Equations

By substitution:
u = (1-x)
du = -dx
∫x/(1-x)dx = -∫(1-u)/u du = ∫1-(1/u)du = u-log(u)+c = (1-x) - log(1-x) + c

By decomposition:
x/(1-x) = 1/(1-x)-1
∫1/(1-x)-1dx = -log(1-x)-x+c

The Attempt at a Solution

Which solutions should I use (1-x) - log(1-x) + c or -log(1-x)-x+c, or are both equally good?

You are integrating, not 'deriving'. But they are both correct. Take the derivative of both and see if they reduce to x/(1-x). They only differ by the constant 1.

 

[greentea]

Sorry, I meant to say "integrate". Thanks for the response though.

 

[nejibanana]

As Dick said, both are correct. A trick that will help you in the future is to combine constants. In this case you have 1 + c. Since both are constants you can combine this into a new constant c' = 1+c. Then since the label of the constant is arbitrary, you can call this new constant c, and both answers are the same. Combining constants works for any operation so two constants multiplied together can be combined into a single constant, etc.