Evaluate the integral: I = integral (x²-x+2)/(x^4-5x²+4)
Evaluate the integral: $$ I = \int \frac{x^2-x+2}{x^4-5x^2+4}\,dx. $$
Explanation of Method
Factoring the denominator (a quadratic in $x^2$) into four distinct linear factors and decomposing into partial fractions reduces the integral to four elementary logarithms.
- Factor $x^4-5x^2+4=(x^2-1)(x^2-4)$ into linear factors.
- Decompose into partial fractions.
- Solve for the coefficients.
- Integrate and combine.
Step 1 · Factor the denominator
$$ x^4-5x^2+4 = (x^2-1)(x^2-4) = (x-1)(x+1)(x-2)(x+2). $$
Step 2 · Set up partial fractions
$$ \frac{x^2-x+2}{(x-2)(x-1)(x+1)(x+2)} = \frac{A}{x-2}+\frac{B}{x-1}+\frac{C}{x+1}+\frac{D}{x+2}. $$
Step 3 · Solve for the coefficients
Using the cover-up method: at $x=2$, $A=\tfrac{4}{1\cdot3\cdot4}=\tfrac13$; at $x=1$, $B=\tfrac{2}{-1\cdot2\cdot3}=-\tfrac13$; at $x=-1$, $C=\tfrac{4}{-3\cdot-2\cdot1}=\tfrac23$; at $x=-2$, $D=\tfrac{8}{-4\cdot-3\cdot-1}=-\tfrac23$.
Step 4 · Integrate and combine
$$ I = \frac13\ln|x-2| - \frac13\ln|x-1| + \frac23\ln|x+1| - \frac23\ln|x+2| + C. $$
Combining logarithms:
$$ I = \frac13\ln\left(\frac{|x-2|(x+1)^2}{|x-1|(x+2)^2}\right) + C. $$
Summary Table
| Coefficient | Value |
|---|---|
| $A$ | $1/3$ |
| $B$ | $-1/3$ |
| $C$ | $2/3$ |
| $D$ | $-2/3$ |
Additional Notes
- The domain excludes $x=\pm1,\pm2$, the four roots of the denominator.
- As a quick sanity check, the four coefficients sum to zero ($\tfrac13-\tfrac13+\tfrac23-\tfrac23=0$), as expected whenever the numerator's degree is at least two less than the denominator's.
- The result can be checked by differentiating the four-log expression: this gives $\tfrac{1}{3(x-2)}-\tfrac{1}{3(x-1)}+\tfrac{2}{3(x+1)}-\tfrac{2}{3(x+2)}$, and combining over the common denominator $(x-2)(x-1)(x+1)(x+2)$ reduces to $\tfrac{x^2-x+2}{x^4-5x^2+4}$, matching the integrand (verified numerically at $x=0$, where both sides equal $\tfrac12$).
Result
The solution to the integral is: