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?

$\int \frac{x-1}{(x+3)(x^2+1)} dx= \int\frac{A}{x+3} + \frac{B(2x) + C}{x^2+1}\,dx $
and
$ x-1= A(x^{2}+1)+ (B(2x)+C)(x+3)$
which is simple. 
