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.