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

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

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By the way, the trick to notice that your integrand is $1+\frac {2x+5}{x^4-1}$ is to "add zero": $$\frac{x^4+2x+4}{x^4-1} = \frac{x^4 -1 + 1+2x+4}{x^4-1} = \frac{x^4 -1}{x^4 - 1} + \frac{1+2x+4}{x^4-1}.$$ –  JavaMan May 7 '13 at 15:45

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.

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I think is CX+D in the last fraction right? –  Ofir Attia May 7 '13 at 15:48
Ahhh...yes...fixed...! ;-) –  amWhy May 7 '13 at 15:49
$2x+5 = x^3(A+B+D)+x^2(B-A+C)+x(A+B-D)+1(B-A-C)$ this is the final equation , now need to find the parameters,thanks! –  Ofir Attia May 7 '13 at 15:57
Actually, you don't need $CX + D$ in the numerator here: it turns out that $C$ is sufficient. I used $$2x + 5 = A(x - 1)(x^2 +1) + B(x+1)(x^2 + 1) + C(x-1)(x+1)$$ and solved for $A$ when $x = -1$, for $B$ when $x = 1$, and for $C$ when $x = 0,$ given $A, B$ –  amWhy May 7 '13 at 16:13
Nice to get the feedback! +1 –  Amzoti May 8 '13 at 0:29

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.

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

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