How $\frac{1}{2a} \log {|\frac{x+a}{x-a}|} + C = \frac{1}{2a} \log {|\frac{a+x}{a-x}|} + C$

Question

I know how $\int{\frac{dx}{x^2-a^2}} = \frac{1}{2a} \log {|\frac{x-a}{x+a}|} + C$

Using the above derivation, I derived
$$\int{\frac{dx}{a^2-x^2}} = -\int{\frac{dx}{x^2-a^2}} = \frac{1}{2a} \log {|\frac{x+a}{x-a}|} + C$$

But my textbook derived differently and got
$$\int{\frac{dx}{a^2-x^2}} = -\int{\frac{dx}{x^2-a^2}} = \frac{1}{2a} \log {|\frac{a+x}{a-x}|} + C$$

Here's how they derived:

How they both are equivalent? Or is my derivation wrong?

Answer 1

They're both equivalent. The absolute value would make it so that the denominator is the same for both answers. i.e. since $x-a = -(a-x)$, $|x-a| = |a-x|$. Absolute value would remove the negative sign.