Integral Of $\frac{x^4+2x+4}{x^4-1}$

Question

Any ideas how to solve it? $$\int\frac{x^4+2x+4}{x^4-1}dx$$ Thanks!

Answer 1

Using polynomial division, we get $$\int \frac{x^4+2x+4}{x^4-1} dx = \int 1 + \frac{2x+5}{(x^2 - 1)(x^2 + 1)}dx = \int 1 + \frac{2x+5}{(x+1)(x-1)(x^2+1)} dx $$

Expressing this as partial fractions, we need only find $A, B, C$

$$= \int \left(1 + \frac{A}{x+1} + \frac B{x-1} +\frac{C}{x^2 + 1}\right)\,dx$$

And then the integration is pretty straightforward.

Answer 2

Hint: First, make the integrand into $1+\frac {2x+5}{x^4-1}$ Now apply partial fractions.

Answer 3

Hint $$ \frac{x^4+2x+4}{x^4-1} = 1 + \frac{2x+5}{x^4-1} = 1 + \frac{2x+5}{(x+1)(x-1)(x^2+1)} $$ and use partial fractions.