Evaluate the integral of (x + 3)/(x² − 5x + 6) with respect to x.

Step-by-step solution

1. 1

   Factor the denominator to choose the right technique.

   $x^2 - 5x + 6 = (x - 2)(x - 3)$

   Two distinct linear factors means the integrand splits into two simple fractions with constant numerators — no repeated-root or irreducible-quadratic complications. The degree of the numerator ($1$) is already below the degree of the denominator ($2$), so no preliminary long division is needed.

2. 2

   Set up the partial-fraction decomposition.

   $\frac{x + 3}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$

   Multiplying through by $(x-2)(x-3)$ clears the denominators:

   $x + 3 = A(x - 3) + B(x - 2)$

3. 3

   Find A and B by substituting the roots. The fastest route is to plug in the values that kill one bracket at a time. Setting $x = 3$:

   $6 = B(3 - 2) \ \Longrightarrow \ B = 6$

   Setting $x = 2$:

   $5 = A(2 - 3) \ \Longrightarrow \ A = -5$

   (Matching coefficients gives the same system $A + B = 1$, $-3A - 2B = 3$.)

4. 4

   Verify the decomposition before integrating. With $A = -5$ and $B = 6$:

   $\frac{-5}{x-2} + \frac{6}{x-3} = \frac{-5(x-3) + 6(x-2)}{(x-2)(x-3)} = \frac{x + 3}{(x-2)(x-3)} \ \checkmark$

   Checking here costs one line and catches sign errors before they propagate into the integral.

5. 5

   Integrate each term. Each piece is of the form $\int \frac{k}{x - a}\,dx = k\ln|x - a|$:

   $\int \left(\frac{-5}{x-2} + \frac{6}{x-3}\right) dx = -5\ln|x - 2| + 6\ln|x - 3| + C$

   The absolute values matter: the antiderivative is valid on each of the three intervals $x < 2$, $2 < x < 3$, $x > 3$ separately.

6. 6

   Differentiate back as a final check.

   $\frac{d}{dx}\left(-5\ln|x-2| + 6\ln|x-3|\right) = \frac{-5}{x-2} + \frac{6}{x-3} = \frac{x+3}{x^2 - 5x + 6} \ \checkmark$

   A numeric spot check at $x = 5$ gives $\frac{8}{6} = 1.3\overline{3}$ from both forms. The answer can also be written as $\ln\left|\frac{(x-3)^6}{(x-2)^5}\right| + C$.