What is the method for finding $\int\frac {x^3 + 5x^2 +2x -4}{(x^4-1)}\mathrm{d}x$? $$
\int\frac {x^3 + 5x^2 +2x -4}{(x^4-1)}dx
$$
A bit confused with how to integrate this question. I though it was partial fractions but was unsure about the what to do after that.

Partial Solution
Well, assuming you did the partial fraction decomposition already (as you said you did), you should get the following integral
$$\int\left(\frac{9-x}{2(x^2+1)}+\frac{1}{x-1}+\frac{1}{2 (x+1)}\right) dx$$
$$=\frac{9}{2}\int\frac {dx}{x^2+1} - \frac{1}{2}\int \frac {x}{x^2+1}dx + \int \frac{dx}{x-1}+ \frac{1}{2}\int \frac{dx}{x+1}$$
A.) $\quad \int\frac {dx}{x^2+1}$ = $\arctan x$
B.) $\quad\int \frac {x}{x^2+1}dx = \frac{1}{2}\int \frac {du}{u+1} = \frac{1}{2}\log (u+1) = \frac{1}{2}\log (x^2+1)$
C.)$\quad\int \frac{dx}{x-1} = \log(x-1)$
D.)$\quad\int \frac{dx}{x+1} = \log(x+1)$
