Evaluate the integral: I = integral dx/(x(x-1)(x+2))

Evaluate the integral: $$ I = \int \frac{dx}{x(x-1)(x+2)}. $$

Step 1 · Set up partial fractions

$$ \frac{1}{x(x-1)(x+2)} = \frac Ax + \frac{B}{x-1} + \frac{C}{x+2}. $$

Multiplying through: $1=A(x-1)(x+2)+Bx(x+2)+Cx(x-1)$.

Step 2 · Solve for the coefficients

Setting $x=0$: $1=A(-1)(2)=-2A$, so $A=-\tfrac12$. Setting $x=1$: $1=B(1)(3)=3B$, so $B=\tfrac13$. Setting $x=-2$: $1=C(-2)(-3)=6C$, so $C=\tfrac16$.

Step 3 · Integrate each term

$$ I = \int \left(\frac{-1/2}{x} + \frac{1/3}{x-1} + \frac{1/6}{x+2}\right) dx = -\frac12\ln|x| + \frac13\ln|x-1| + \frac16\ln|x+2| + C. $$

Summary Table

Additional Notes

• The domain excludes $x=0,1,-2$, the three roots of the denominator.
• As a quick sanity check, the coefficients $A,B,C$ always sum to $0$ when the denominator's degree exceeds the numerator's degree by at least $2$ (here, $3-0=3\geq2$): $-\tfrac12+\tfrac13+\tfrac16=-\tfrac36+\tfrac26+\tfrac16=0$, confirming no arithmetic slip.
• The result can be checked by differentiating $-\tfrac12\ln|x|+\tfrac13\ln|x-1|+\tfrac16\ln|x+2|$: this gives $-\tfrac{1}{2x}+\tfrac{1}{3(x-1)}+\tfrac{1}{6(x+2)}$, which combines over $6x(x-1)(x+2)$ to numerator $-3(x-1)(x+2)+2x(x+2)+x(x-1)=6$, reducing to $\tfrac{6}{6x(x-1)(x+2)}=\tfrac{1}{x(x-1)(x+2)}$, matching the integrand.