Evaluate the integral: I = integral (x^6)/(x²-1) dx

Evaluate the integral: $$ I = \int \frac{x^6}{x^2-1}\,dx. $$

Step 1 · Polynomial long division

$$ \frac{x^6}{x^2-1} = x^4+x^2+1+\frac{1}{x^2-1}. $$

Step 2 · Integrate the polynomial part

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

Step 3 · Integrate the remainder

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

$$ \int \frac{dx}{x^2-1} = \frac12\ln\left|\frac{x-1}{x+1}\right| + C. $$

Step 4 · Combine

$$ I = \frac{x^5}{5}+\frac{x^3}{3}+x + \frac12\ln\left|\frac{x-1}{x+1}\right| + C. $$

Summary Table

Additional Notes

• The domain excludes $x=\pm1$, the roots of the denominator.
• The polynomial quotient $x^4+x^2+1$ pattern-matches $x^6-1=(x^2-1)(x^4+x^2+1)$, so this division could also be read off directly from the sum-of-geometric-series identity $x^6-1=(x^2)^3-1$.
• The result can be checked by differentiating $\tfrac{x^5}5+\tfrac{x^3}3+x+\tfrac12\ln\left|\tfrac{x-1}{x+1}\right|$: this gives $x^4+x^2+1+\tfrac{1}{x^2-1}=\tfrac{(x^4+x^2+1)(x^2-1)+1}{x^2-1}=\tfrac{x^6-1+1}{x^2-1}=\tfrac{x^6}{x^2-1}$, matching the integrand.