Evaluate the integral: I = integral (x-4)/(x²-5x+6) dx

Evaluate the integral: $$ I = \int \frac{x-4}{x^2-5x+6}\,dx. $$

Step 1 · Factor and set up partial fractions

$$ x^2-5x+6 = (x-2)(x-3), \qquad \frac{x-4}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}. $$

Multiplying through: $x-4=A(x-3)+B(x-2)$. Setting $x=2$: $-2=-A$, so $A=2$. Setting $x=3$: $-1=B$.

Step 2 · Integrate

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

Step 3 · Combine logarithms

$$ I = \ln\left|\frac{(x-2)^2}{x-3}\right| + C. $$

Summary Table

Additional Notes

• The domain excludes $x=2$ and $x=3$, the roots of the denominator.
• This follows the same cover-up-method pattern as Problem 210, just with both denominator roots positive.
• The result can be checked by differentiating $2\ln|x-2|-\ln|x-3|$: this gives $\tfrac{2}{x-2}-\tfrac{1}{x-3}=\tfrac{2(x-3)-(x-2)}{(x-2)(x-3)}=\tfrac{x-4}{(x-2)(x-3)}$, matching the integrand.