Evaluate the integral: I = integral x/(x^8-1) dx
Evaluate the integral: $$ I = \int \frac{x}{x^8-1}\,dx. $$
Explanation of Method
Substituting $u=x^2$ reduces the octic denominator to a quartic in $u$, which factors into linear and quadratic pieces; partial fractions then finish the job.
- Substitute $u=x^2$.
- Factor $u^4-1$.
- Decompose into partial fractions.
- Integrate and substitute back.
Step 1 · Substitute
Let $u=x^2$, $du=2x\,dx$:
$$ I = \frac12\int \frac{du}{u^4-1}. $$
Step 2 · Factor and set up partial fractions
$$ u^4-1=(u-1)(u+1)(u^2+1), \qquad \frac{1}{(u-1)(u+1)(u^2+1)} = \frac{A}{u-1}+\frac{B}{u+1}+\frac{Cu+D}{u^2+1}. $$
Solving gives $A=\tfrac14$, $B=-\tfrac14$, $C=0$, $D=-\tfrac12$.
Step 3 · Integrate
$$ \frac12\int \left(\frac{1/4}{u-1}-\frac{1/4}{u+1}-\frac{1/2}{u^2+1}\right) du = \frac18\ln\left|\frac{u-1}{u+1}\right| - \frac14\arctan u + C. $$
Step 4 · Substitute back
With $u=x^2$:
$$ I = \frac18\ln\left|\frac{x^2-1}{x^2+1}\right| - \frac14\arctan\left(x^2\right) + C. $$
Summary Table
| Coefficient | Value |
|---|---|
| $A$ | $1/4$ |
| $B$ | $-1/4$ |
| $C$ | $0$ |
| $D$ | $-1/2$ |
Additional Notes
- The domain excludes $x=\pm1$, where the denominator vanishes.
- The substitution $u=x^2$ is the key move here: since only $x\,dx$ (not a lone $dx$) appears, the odd power of $x$ in the numerator matches exactly with what $du=2x\,dx$ needs, converting an octic-denominator integral into a much more tractable quartic one.
- The result can be checked by differentiating $\tfrac18\ln\left|\tfrac{x^2-1}{x^2+1}\right|-\tfrac14\arctan(x^2)$: the log term's derivative is $\tfrac{4x}{8(x^4-1)}=\tfrac{x}{2(x^4-1)}$ (using $\tfrac{d}{dx}\ln|(x^2-1)/(x^2+1)|=4x/(x^4-1)$), and the arctangent term's derivative is $-\tfrac{x}{2(1+x^4)}$; at $x=2$ these evaluate to $\tfrac{2}{2(15)}=\tfrac1{15}$ and $-\tfrac{2}{2(17)}=-\tfrac1{17}$, summing to $\tfrac1{15}-\tfrac1{17}=\tfrac2{255}$, matching the integrand $\tfrac{2}{2^8-1}=\tfrac{2}{255}$.
Result
The solution to the integral is: