Evaluate the integral: I = integral dx/(x^4-1)

Evaluate the integral: $$ I = \int \frac{dx}{x^4-1}. $$

Step 1 · Factor and set up partial fractions

$$ x^4-1=(x^2-1)(x^2+1), \qquad \frac{1}{(x^2-1)(x^2+1)} = \frac{A}{x^2-1} + \frac{B}{x^2+1}. $$

Multiplying through: $1=A(x^2+1)+B(x^2-1)$. Matching coefficients: $A+B=0$ and $A-B=1$, giving $A=\tfrac12$, $B=-\tfrac12$.

Step 2 · Decompose $1/(x^2-1)$ further

$$ \frac{1}{x^2-1} = \frac12\left(\frac{1}{x-1}-\frac{1}{x+1}\right). $$

Step 3 · Integrate each piece

$$ \frac12\int \frac{dx}{x^2-1} = \frac14\ln|x-1| - \frac14\ln|x+1|, \qquad -\frac12\int \frac{dx}{x^2+1} = -\frac12\arctan x. $$

Step 4 · Combine

$$ I = \frac14\ln\left|\frac{x-1}{x+1}\right| - \frac12\arctan x + C. $$

Summary Table

Additional Notes

• The domain excludes $x=\pm1$, the real roots of $x^4-1$.
• This two-level decomposition (first splitting $x^4-1$ into $x^2-1$ and $x^2+1$, then splitting $x^2-1$ further into linear factors) is more efficient here than attempting a single four-term partial fraction decomposition directly.
• The result can be checked by differentiating $\tfrac14\ln|x-1|-\tfrac14\ln|x+1|-\tfrac12\arctan x$: this gives $\tfrac1{4(x-1)}-\tfrac1{4(x+1)}-\tfrac1{2(1+x^2)} = \tfrac{1}{2(x^2-1)}-\tfrac1{2(x^2+1)} = \tfrac{(x^2+1)-(x^2-1)}{2(x^4-1)} = \tfrac{2}{2(x^4-1)} = \tfrac{1}{x^4-1}$, matching the integrand.