Evaluate the integral: I = integral (3-2x)/(x²-x-2) dx

Evaluate the integral: $$ I = \int \frac{3-2x}{x^2-x-2}\,dx. $$

Step 1 · Factor and set up partial fractions

$$ x^2-x-2 = (x+1)(x-2), \qquad \frac{3-2x}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}. $$

Multiplying through: $3-2x=A(x-2)+B(x+1)$. Setting $x=-1$: $5=-3A$, so $A=-\tfrac53$. Setting $x=2$: $-1=3B$, so $B=-\tfrac13$.

Step 2 · Integrate

$$ I = \int \left(\frac{-5/3}{x+1} + \frac{-1/3}{x-2}\right) dx = -\frac53\ln|x+1| - \frac13\ln|x-2| + C. $$

Summary Table

Additional Notes

• The domain excludes $x=-1$ and $x=2$, the roots of the denominator.
• Both partial-fraction coefficients here are negative, which is easy to double-check via the cover-up method: plugging $x=-1$ into $(3-2x)/(x-2)$ gives $5/(-3)=-5/3=A$, and plugging $x=2$ into $(3-2x)/(x+1)$ gives $-1/3=B$, matching the solved system.
• The result can be checked by differentiating $-\tfrac53\ln|x+1|-\tfrac13\ln|x-2|$: this gives $-\tfrac{5}{3(x+1)}-\tfrac{1}{3(x-2)}=\tfrac{-5(x-2)-(x+1)}{3(x+1)(x-2)}=\tfrac{-6x+9}{3(x^2-x-2)}=\tfrac{3-2x}{x^2-x-2}$, matching the integrand.