Solving Integration by Parts Problem: \int\frac{dx}{a^2-x^2}

[mbrmbrg]

Problem:
\int\frac{dx}{a^2-x^2}
My Work:
\frac{1}{a^2-x^2}
=\frac{1}{(a+x)(a-x)}=\frac{A}{a+x}+\frac{B}{a-x}}
1=A(a-x)+B(a+x)
If x=a, then 1=2Ba so B=\frac{1}{2a}
Thus 1=A(a-x)+\frac{1}{2a}(a+x)
if x=0, then 1=Aa+\frac{1}{2} so A=\frac{1}{2a}
SO
\int\frac{dx}{a^2-x^2}
=\int\frac{\frac{1}{2a}}{a+x}+\frac{\frac{1}{2a}}{a-x}dx
=\frac{1}{2a}\int\frac{dx}{a+x}+\frac{1}{2a}\int\frac{dx}{a-x}
=\frac{1}{2a}\ln\mid a+x\mid +\frac{1}{2a}\ln\mid a-x\mid(-1)
=\frac{1}{2a}(\ln\mid a+x\mid -\ln\mid a-x\mid)
=\frac{1}{2a}\ln\mid\frac{a+x}{a-x}\mid +C

Table of integrals gives correct answer as
=\frac{1}{2a}\ln\mid\frac{x+a}{x-a}\mid +C

My gut feeling is that I messed up integrating \int\frac{dx}{a-x} but I can't find my error.
Any help would be appreciated.

 

[courtrigrad]

Its the same thing, because its in absolute values.

 

[mbrmbrg]

Thanks, Courtrigrad.