Evaluate the integral: I = integral (x^8-1)/(x (x^8+1)) dx

Evaluate the integral: $$ I = \int \frac{x^8-1}{x\left(x^8+1\right)}\,dx. $$

Step 1 · Rewrite and split

$$ \frac{x^8-1}{x(x^8+1)} = \frac{(x^8+1)-2}{x(x^8+1)} = \frac1x - \frac{2}{x(x^8+1)}. $$

Using the identity $\dfrac{1}{x(x^8+1)}=\dfrac1x-\dfrac{x^7}{x^8+1}$ (since $(x^8+1)-x^8=1$):

$$ \frac{x^8-1}{x(x^8+1)} = \frac1x - 2\left(\frac1x - \frac{x^7}{x^8+1}\right) = -\frac1x + \frac{2x^7}{x^8+1}. $$

Step 2 · Integrate each piece

Let $w=x^8+1$, $dw=8x^7\,dx$:

$$ I = -\int \frac{dx}{x} + 2\int \frac{x^7}{x^8+1}\,dx = -\ln|x| + \frac14\ln\left(x^8+1\right) + C. $$

Summary Table

Additional Notes

• The domain excludes $x=0$, where the denominator vanishes.
• The identity $1/[x(x^8+1)]=1/x-x^7/(x^8+1)$ is the same "add and subtract" trick used in Problem 255, generalized to an 8th-power denominator; it avoids a full 9-term partial-fraction decomposition of $x(x^8+1)$ into all its linear and quadratic factors.
• The result can be checked by differentiating $-\ln|x|+\tfrac14\ln(x^8+1)$: this gives $-\tfrac1x+\tfrac{2x^7}{x^8+1}$, and combining over $x(x^8+1)$ gives numerator $-(x^8+1)+2x^8=x^8-1$, matching the integrand exactly.