Integration Question: Completing the Square/Trig Sub yields a different answer than integral table. After the completing the square,
$$\int \frac{dx}{x^2 + 2x - 3}$$
becomes,
$$ \int \frac{dx}{(x+1)^2 - 4}$$
The integral table in my book says the antiderivative is,
$$\frac{1}{2a} ln \, \Biggl\lvert \frac{x-a}{x+a} \Biggr\rvert + \, C$$
or,
$$ \frac{1}{4} ln \, \Biggl\lvert \frac{x-1}{x+3} \Biggr\rvert + \, C$$
Using trig sub, and $\displaystyle u=a\sec\theta$ I obtained,
$$\frac{1}{2}\int csc \, \theta \, d\theta$$ 
which corresponds to
$$-\frac{1}{2} ln \, \Biggl\lvert \frac{x+3}{\sqrt{x^2+2x-3}} \Biggr\rvert + \, C $$
I am having a lot of trouble putting this result in the form presented in the table. Any help would be much appreciated.

$$-\frac{1}{2} \ln \, \Biggl\lvert \frac{x+3}{\sqrt{x^2+2x-3}} \Biggr\rvert = -\frac12 \ln \left| \sqrt{\frac{(x+3)^2}{(x-1)(x+3)}} \right| = -\frac14 \ln \left| \frac{x+3}{x-1} \right| = \frac14 \ln \left|\frac{x-1}{x+3}\right|$$
