Disclaimer: I am not a student trying to get free internet homework help. I am an adult who is learning Calculus. I am deeply grateful to the members of this community for their time.
$$\int{\frac{1}{(x^2+2x-4)(x+3)}}dx$$
This led me to $$A=1, B=-1, C=-1$$
However I don't know how to integrate the following: $$\int{\frac{x-1}{(x^2+2x-4)}}dx$$
Here is my work:
$$ \begin{split} \int \frac{x-1}{(x^2+2x-4)} dx &= \int \frac{x+1}{x^2+2x-4} + \int \frac{-2}{x^2+2x-4} \\ &= \frac{1}{2} \ln\left|x^2+2x-4\right| + (-2)\int\frac{dx}{(x+1)^2-5} \end{split} $$
$$\int \frac{dx}{(x+1)^2-5}=\frac{1}{2\sqrt{5}}\int(\frac{1}{x+1-\sqrt{5}}-\frac{1}{x+1+\sqrt{5}})dx $$