Evaluate the integral: I = integral dx/(x^4-x²-2)
Evaluate the integral: $$ I = \int \frac{dx}{x^4-x^2-2}. $$
Explanation of Method
Factoring the denominator (a quadratic in $x^2$) into an irreducible quadratic times a difference of squares, then decomposing into partial fractions, reduces the integral to a logarithm plus an arctangent.
- Factor $x^4-x^2-2=(x^2-2)(x^2+1)$.
- Decompose into partial fractions.
- Solve for the coefficients.
- Integrate each term.
Step 1 · Factor and set up partial fractions
$$ x^4-x^2-2 = (x^2-2)(x^2+1), \qquad \frac{1}{(x^2-2)(x^2+1)} = \frac{Ax+B}{x^2-2} + \frac{Cx+D}{x^2+1}. $$
Step 2 · Solve for the coefficients
Multiplying through and matching coefficients gives $A=C=0$ (by even symmetry), and solving the remaining system yields $B=\tfrac13$, $D=-\tfrac13$.
Step 3 · Integrate each term
$$ \frac{1}{(x^2-2)(x^2+1)} = \frac{1/3}{x^2-2} - \frac{1/3}{x^2+1}. $$
Using the standard results $\int dx/(x^2-a^2)=\tfrac1{2a}\ln\left|\tfrac{x-a}{x+a}\right|$ with $a=\sqrt2$, and $\int dx/(x^2+1)=\arctan x$:
$$ I = \frac13\cdot\frac{1}{2\sqrt2}\ln\left|\frac{x-\sqrt2}{x+\sqrt2}\right| - \frac13\arctan x + C. $$
Summary Table
| Coefficient | Value |
|---|---|
| $A,C$ | $0$ |
| $B$ | $1/3$ |
| $D$ | $-1/3$ |
Additional Notes
- The domain excludes $x=\pm\sqrt2$, where the denominator vanishes.
- Since $x^4-x^2-2$ is an even function, its reciprocal's decomposition contains only even-symmetric terms, ruling out $A$ and $C$ from the start.
- The result can be checked by differentiating $\tfrac{1}{6\sqrt2}\ln\left|\tfrac{x-\sqrt2}{x+\sqrt2}\right|-\tfrac13\arctan x$: this gives $\tfrac{1}{3(x^2-2)}-\tfrac{1}{3(x^2+1)}$, and combining over $3(x^2-2)(x^2+1)$ gives numerator $(x^2+1)-(x^2-2)=3$, reducing to $\tfrac{1}{(x^2-2)(x^2+1)}$, matching the integrand.
Result
The solution to the integral is: