Very quick and easy question about integrating and completing the square

[twisted079]

Homework Statement

The problem could be any variation of dx/(x²-2x)

Homework Equations

∫dx/x²-a² = 1/2a ln ((x-a)/(x+a)) + C

The Attempt at a Solution

I understand the answer to be 1/2 ln ((x-2)/x) + C

My question is why is it just x on the bottom in the solution? Shouldnt it be x+2 since the equation states "x+a"? Similar to the way it is x-2 on the top which makes sense since the formula states "x-a". Any help is greatly appreciated.

 

[Mark44]

Homework Statement

The problem could be any variation of dx/(x²-2x)

Homework Equations

∫dx/x²-a² = 1/2a ln ((x-a)/(x+a)) + C

This integration formula isn't directly relevant to your problem. In this formula, the denominator is a difference of squares. In your problem, you don't have a difference of squares.

You could manipulate your denominator to get a difference of squares, since
x² - 2x = x² - 2x + 1 - 1 = (x - 1)² = 1.

The Attempt at a Solution

I understand the answer to be 1/2 ln ((x-2)/x) + C

My question is why is it just x on the bottom in the solution? Shouldnt it be x+2 since the equation states "x+a"? Similar to the way it is x-2 on the top which makes sense since the formula states "x-a". Any help is greatly appreciated.

 

[twisted079]

Ah yes, I overlooked the minor details. Thank you.