What is the integral of $\frac{x-1}{(x+3)(x^2+1)}$? I've worked with partial fractions to get the integral in the form $$\int\frac{A}{x+3} + \frac{Bx + C}{x^2+1}\,dx$$ Is there a quicker way?

First of all: you have to be carefull with your notation because you mean:
$$\int\left(\frac{A}{x+3} + \frac{Bx + C}{x^2+1}\right)\space\text{d}x$$

And Partial fractions is the most easy way:
$$\int\frac{x-1}{(x+3)(x^2+1)}\space\text{d}x=$$
$$\int\left(\frac{2x-1}{5(x^2+1)}-\frac{2}{5(x+3)}\right)\space\text{d}x=$$
$$\frac{1}{5}\int\frac{2x-1}{x^2+1}\space\text{d}x-\frac{2}{5}\int\frac{1}{x+3}\space\text{d}x=$$
$$\frac{1}{5}\int\left(\frac{2x}{x^2+1}-\frac{1}{x^2+1}\right)\space\text{d}x-\frac{2}{5}\int\frac{1}{x+3}\space\text{d}x$$
