Evaluate the integral: I = integral x/(x³-3x+2) dx
Evaluate the integral: $$ I = \int \frac{x}{x^3-3x+2}\,dx. $$
Explanation of Method
Testing rational roots reveals a repeated factor in the denominator; decomposing into partial fractions over a linear factor, its square, and a distinct linear factor reduces the integral to logarithms and a power-rule term.
- Factor the denominator using the rational root theorem.
- Decompose into partial fractions.
- Solve for the coefficients.
- Integrate each term.
Step 1 · Factor the denominator
Testing $x=1$: $1-3+2=0$, so $(x-1)$ is a factor. Dividing gives $x^3-3x+2=(x-1)(x^2+x-2)=(x-1)(x-1)(x+2)=(x-1)^2(x+2)$.
Step 2 · Set up partial fractions
$$ \frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}. $$
Multiplying through: $x=A(x-1)(x+2)+B(x+2)+C(x-1)^2$.
Step 3 · Solve for the coefficients
Setting $x=1$: $1=3B$, so $B=\tfrac13$. Setting $x=-2$: $-2=9C$, so $C=-\tfrac29$. Setting $x=0$ and using $B,C$: $0=-2A+2B+C=-2A+\tfrac23-\tfrac29$, giving $A=\tfrac29$.
Step 4 · Integrate each term
$$ I = \int \left(\frac{2/9}{x-1} + \frac{1/3}{(x-1)^2} - \frac{2/9}{x+2}\right) dx = \frac29\ln|x-1| - \frac{1}{3(x-1)} - \frac29\ln|x+2| + C. $$
Summary Table
| Coefficient | Value |
|---|---|
| $A$ | $2/9$ |
| $B$ | $1/3$ |
| $C$ | $-2/9$ |
Additional Notes
- The domain excludes $x=1$ and $x=-2$, the roots of the denominator.
- The rational root theorem narrows candidate roots of $x^3-3x+2$ to divisors of the constant term ($\pm1,\pm2$), making $x=1$ quick to find by direct substitution before any long division is needed.
- The result can be checked by differentiating $\tfrac29\ln|x-1|-\tfrac1{3(x-1)}-\tfrac29\ln|x+2|$: this gives $\tfrac{2}{9(x-1)}+\tfrac{1}{3(x-1)^2}-\tfrac{2}{9(x+2)}$, matching the partial-fraction decomposition (and hence the integrand) term by term.
Result
The solution to the integral is: