# Solving the Integral ∫dx/(1-x)

## Homework Statement:

Solve the following integral

## Relevant Equations:

∫dx/(1-x)
I solved the integral by two different methods and I get different answers.

Method 1:
∫dx/(1-x) = -∫-dx/(1-x), u=1-x, du=-dx

∫dx/(1-x) = -∫du/u = -ln|u| = -ln|1-x|

Method 2:
∫-dx/(x-1) = -∫dx/(x-1), u=x-1, du=dx

∫-dx/(x-1) = -∫du/u = -ln|u| = -ln|x-1|

What am I doing wrong?

• Delta2

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vela
Staff Emeritus
Homework Helper
Try calculating both quantities for, say, $x=3$.

Mark44
Mentor
Homework Statement:: Solve the following integral
Relevant Equations:: ∫dx/(1-x)

I solved the integral by two different methods and I get different answers.

Method 1:
∫dx/(1-x) = -∫-dx/(1-x), u=1-x, du=-dx

∫dx/(1-x) = -∫du/u = -ln|u| = -ln|1-x|

Method 2:
∫-dx/(x-1) = -∫dx/(x-1), u=x-1, du=dx

∫-dx/(x-1) = -∫du/u = -ln|u| = -ln|x-1|

What am I doing wrong?
At heart, your question really isn't about calculus -- it's about a property of the absolute value.
|a - b| = |b - a|, right?

• SammyS, archaic and Delta2
What am I doing wrong?
Nothing