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.

$$-\frac{1}{2} ln \, \Biggl\lvert \frac{x+3}{\sqrt{x^2+2x-3}} \Biggr\rvert + \, C $$
You can take the square of the term in the absolut value signs. At the same time you take the square root. Remember the rule $log|x^a|=a\cdot log|x|$
$$-\frac{1}{2} ln \, \Biggl\lvert \left( \frac{(x+3)^2}{x^2+2x-3 } \right) ^{1/2}\Biggr\rvert + \, C=-\frac{1}{4} ln \, \Biggl\lvert  \frac{(x+3)^2}{x^2+2x-3 } \Biggr\rvert + \, C $$ 
Linear factors of the denominator
$$=-\frac{1}{4} ln \, \Biggl\lvert \frac{(x+3)^2}{(x+3)\cdot(x-1)   } \Biggr\rvert + \, C$$
Cancelling $(x+3)$. And taking away the negative sign. To neutralize this change you have to exchange the numerator and denominator.
$$=\frac{1}{4} ln \, \Biggl\lvert \frac{x-1}{x+3   }  \Biggr\rvert + \, C$$
