Evaluate the integral: I = integral dx/(x²(x^4-1))
Evaluate the integral: $$ I = \int \frac{dx}{x^2(x^4-1)}. $$
Explanation of Method
Factoring $x^4-1$ completely into linear and quadratic factors and decomposing into partial fractions reduces the integral to a power-rule term, two logarithms, and an arctangent.
- Factor $x^4-1=(x-1)(x+1)(x^2+1)$.
- Decompose into partial fractions.
- Solve for the coefficients.
- Integrate each term.
Step 1 · Factor and set up partial fractions
$$ \frac{1}{x^2(x-1)(x+1)(x^2+1)} = \frac Ax + \frac{B}{x^2} + \frac{C}{x-1} + \frac{D}{x+1} + \frac{E}{x^2+1}. $$
Step 2 · Solve for the coefficients
Setting $x=1$: $C=\tfrac14$. Setting $x=-1$: $D=-\tfrac14$. Multiplying by $x^2$ and setting $x=0$: $B=-1$. Multiplying by $x^2+1$ and setting $x^2=-1$: $E=\tfrac12$. Matching the remaining coefficient gives $A=0$.
Step 3 · Integrate each term
$$ \frac{1}{x^2(x^4-1)} = \frac{-1}{x^2} + \frac{1/4}{x-1} - \frac{1/4}{x+1} + \frac{1/2}{x^2+1}. $$
$$ I = \frac1x + \frac14\ln|x-1| - \frac14\ln|x+1| + \frac12\arctan x + C. $$
Summary Table
| Coefficient | Value |
|---|---|
| $A$ | $0$ |
| $B$ | $-1$ |
| $C$ | $1/4$ |
| $D$ | $-1/4$ |
| $E$ | $1/2$ |
Additional Notes
- The domain excludes $x=0,\pm1$, the roots of the denominator.
- The coefficient $A=0$ can be predicted in advance: since $1/[x^2(x^4-1)]$ is an even function, its partial-fraction decomposition can only contain even-symmetric terms ($1/x^2$, and the $1/(x-1)-1/(x+1)$ pairing, and $1/(x^2+1)$), never a lone $1/x$ term.
- The result can be checked by differentiating $\tfrac1x+\tfrac14\ln|x-1|-\tfrac14\ln|x+1|+\tfrac12\arctan x$: this gives $-\tfrac1{x^2}+\tfrac{1}{4(x-1)}-\tfrac1{4(x+1)}+\tfrac{1}{2(x^2+1)}$, matching the partial-fraction decomposition (and hence the integrand) term by term.
Result
The solution to the integral is: