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?

Notice, for $(Bx+C)$ part, you should use separation as follows  $$\int \frac{x-1}{(x+3)(x^2+1)}\ dx=\int \frac{-2}{5(x+3)}+\frac{2x-1}{5(x^2+1)}\ dx$$
$$=\frac{1}{5}\int \left(-\frac{2}{x+3}+\frac{2x}{x^2+1}-\frac{1}{x^2+1}\right)\ dx$$
$$=\frac{1}{5} \left(-2\int \frac{1}{x+3}\ dx+\int \frac{d(x^2)}{x^2+1}-\int\frac{1}{x^2+1}\ dx\right)$$
