How can I solve this primitive function?

Question

The primitive function I'm trying to solve.

$\int_\frac{1}{x^4-1}\;dx$

• I've used partial fraction decomposition method.
• The following is the equation I'm setting up to be able to solve A, B and C:$(1/(x^2-1)(x^2+1)) = A/(x^2-1)+(Bx+C)/(x^2+1) $
  • $1 = A(x^2+1) + (Bx+C)(x^2-1)$
  • $1 = Ax^2 + A + Bx^3-Bx+Cx^2-C$
  • $A = 1/2 , B = 0, C = -(1/2)$
• I end up getting $\frac 12 *ln|x^2-1|-\frac 12arctan (x)+ C$ as my final answer every time.

What am I missing?

Answer 1

$$ \begin{aligned} \because \frac{1}{x^{4}-1} &=\frac{1}{2}\left(\frac{1}{x^{2}-1}-\frac{1}{x^{2}+1}\right) \\ &=\frac{1}{2}\left[\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)-\frac{1}{x^{2}+1}\right] \\ \therefore \int \frac{1}{x^{4}-1} d x &=\frac{1}{4}\left[\int \frac{d x}{x-1}-\int \frac{d x}{x+1}\right]-\frac{1}{2} \int \frac{d x}{x^{2}+1} \\ &=\frac{1}{4} \ln \left|\frac{x-1}{x+1}\right|-\frac{1}{2} \tan ^{-1} x+C \end{aligned} $$