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

• greentea
In summary, there are two methods for integrating x/(1-x), substitution and decomposition, both resulting in the same answer of (1-x) - log(1-x) + c or -log(1-x)-x+c. The difference between the two is a constant, but this can be combined and both answers are equally valid.
greentea

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?

greentea said:

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.

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

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.

## 1. What is substitution and decomposition in relation to solving equations?

Substitution and decomposition are algebraic techniques used to solve equations. Substitution involves replacing a variable in an equation with a value or expression, while decomposition involves breaking down a complex equation into simpler parts.

## 2. How do you use substitution to solve an equation?

To use substitution, you first identify a variable in the equation that you want to eliminate. Then, you solve for that variable in terms of the other variables in the equation. Finally, you substitute the resulting expression into the original equation and solve for the remaining variable.

## 3. When is decomposition helpful in solving equations?

Decomposition is helpful for solving equations that contain multiple terms or are raised to a power. By breaking down the equation into simpler parts, it becomes easier to solve for the variable.

## 4. Can you use both substitution and decomposition together to solve an equation?

Yes, it is possible to use both substitution and decomposition together to solve an equation. In some cases, using both techniques may be necessary to fully simplify and solve the equation.

## 5. Are there any limitations to using substitution and decomposition to solve equations?

Substitution and decomposition may not work for every equation. Some equations may require more advanced techniques or may not have a solution that can be expressed algebraically.

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