What is the method for finding $\int\frac {x^3 + 5x^2 +2x -4}{(x^4-1)}\mathrm{d}x$?

Question

$$ \int\frac {x^3 + 5x^2 +2x -4}{(x^4-1)}dx $$ 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.

Answer 1

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.)$\quad\int \frac{dx}{x+1} = \log(x+1)$

Answer 2

The integral can be written as

$$I=\frac{1}{4}\int\frac{4x^3}{x^4-1}+5\int\frac{x^2}{(x^2-1)(x^2+1)}+\int\frac{2x}{(x^2-1)(x^2+1)}-4\int\frac{dx}{(x^2-1)(x^2+1)}=I_1+I_2+I_3+I_4$$

$$I_1=\frac{1}{4}ln|x^4-1|+C$$

$$I_2=\frac{5}{2}\int\frac{1}{x^2-1}+\int\frac{1}{x^2+1}$$

$$I_3=\int\frac{dt}{(t-1)(t+1)}$$ where $t=x^2$

$$I_4=-2\int\frac{1}{x^2-1}-2\int\frac{1}{x^2+1}$$