<s>[INST] 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.
 [/INST] 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.)$\quad\int \frac{dx}{x+1} = \log(x+1)$
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