Integrate the function: $\frac{x+3}{x^{2}-2x-5}$

Let $(x+3) = A \frac{d}{dx}(x^{2}-2x-5) + B$
$(x+3) = A(2x-2) + B$
Equating the coefficients of $x$ and the constant term on both sides,we obtain:
$2A = 1 \Rightarrow A = \frac{1}{2}$
$-2A + B = 3 \Rightarrow -2(\frac{1}{2}) + B = 3 \Rightarrow -1 + B = 3 \Rightarrow B = 4$
Therefore,$(x+3) = \frac{1}{2}(2x-2) + 4$
Now,$\int \frac{x+3}{x^{2}-2x-5} dx = \int \frac{\frac{1}{2}(2x-2)+4}{x^{2}-2x-5} dx$
$= \frac{1}{2} \int \frac{2x-2}{x^{2}-2x-5} dx + 4 \int \frac{1}{x^{2}-2x-5} dx$
Let $I_{1} = \int \frac{2x-2}{x^{2}-2x-5} dx$ and $I_{2} = \int \frac{1}{x^{2}-2x-5} dx$
For $I_{1}$,let $x^{2}-2x-5 = t$,then $(2x-2)dx = dt$. So,$I_{1} = \int \frac{dt}{t} = \log |x^{2}-2x-5|$.
For $I_{2}$,$x^{2}-2x-5 = (x-1)^{2} - 6 = (x-1)^{2} - (\sqrt{6})^{2}$.
Using $\int \frac{1}{x^{2}-a^{2}} dx = \frac{1}{2a} \log |\frac{x-a}{x+a}|$,we get $I_{2} = \frac{1}{2\sqrt{6}} \log |\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|$.
Substituting $I_{1}$ and $I_{2}$ back,the integral is $\frac{1}{2} \log |x^{2}-2x-5| + \frac{4}{2\sqrt{6}} \log |\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}| + C$
$= \frac{1}{2} \log |x^{2}-2x-5| + \frac{2}{\sqrt{6}} \log |\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}| + C$.