Example 12: $\int\frac{x^2+1}{x^2-5x+6}\,dx$

Basics you need

Solution

1. Factorise the bottom and note where the integrand lives

   We need two numbers with product $6$ and sum $5$⁠: they are $2$ and $3$⁠. So

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

   The integrand is therefore defined for all real $x$ except $x = 2$ and $x = 3$⁠.

   Partial fractions are built from the factors of the bottom, so factorising is always the first job. Knowing where the bottom is zero also tells us in advance where the final answer can be valid: nowhere near $x = 2$ or $x = 3$ can the integral be taken across.
   Check the factorisation by expanding: $(x-2)(x-3)$ is $x^2 - 3x - 2x + 6$⁠, which is $x^2 - 5x + 6$⁠. Basics: splitting fractions

2. The fraction is improper, so divide first

   The top $x^2 + 1$ and the bottom $x^2 - 5x + 6$ both have degree 2, so the fraction is not proper. Dividing $x^2+1$ by $x^2-5x+6$⁠: the quotient is $1$⁠, and the remainder is

   $$(x^2+1) - 1\cdot(x^2-5x+6) = 5x - 5.$$

   So

   $$\frac{x^2+1}{x^2-5x+6} = 1 + \frac{5x-5}{(x-2)(x-3)}.$$
   Table 7.2 only applies to proper fractions. If we tried $x^2 + 1 = A(x-3) + B(x-2)$ directly, the left side has an $x^2$ term and the right side cannot produce one, so no $A$⁠, $B$ would work. Division removes the whole-number part $1$ and leaves a remainder of degree 1, lower than the degree 2 of the bottom.
   Dividend = quotient × divisor + remainder, with the remainder of lower degree than the divisor. Here $x^2 + 1 = 1\cdot(x^2-5x+6) + (5x-5)$⁠. Lesson: partial fractions

3. Set up the partial fractions

   The bottom has two different linear factors, so by Table 7.2, row 1, we write

   $$\frac{5x-5}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}.$$

   Multiply both sides by $(x-2)(x-3)$⁠:

   $$5x - 5 = A(x-3) + B(x-2).$$
   Multiplying through clears every denominator and leaves an equation between two polynomials. It has to hold for every $x$ (it is an identity, not an equation to solve), and that is what lets us find $A$ and $B$ in the next step.
   Two polynomials that are equal for every $x$ have equal coefficients of each power of $x$⁠. Basics: splitting fractions

4. Find $A$ and $B$

   Expand the right side:

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

   Equating the coefficients of $x$ and the constant terms with those of $5x - 5$⁠:

   $$A + B = 5, \qquad 3A + 2B = 5.$$

   From the first, $B = 5 - A$⁠. Then $3A + 2(5 - A) = 5$⁠, so $A + 10 = 5$ and

   $$A = -5, \qquad B = 10.$$

   Quick cross-check. Put $x = 2$ in $5x - 5 = A(x-3) + B(x-2)$⁠: $5 = -A$⁠, so $A = -5$⁠. Put $x = 3$⁠: $10 = B$⁠. The same values. Hence

   $$\frac{x^2+1}{x^2-5x+6} = 1 - \frac{5}{x-2} + \frac{10}{x-3}.$$
   Comparing coefficients is the book's method. Substituting the roots $x = 2$ and $x = 3$ is a fast independent check: each choice makes one bracket zero and leaves a single unknown. (This is legitimate even though the original fraction is undefined at 2 and 3: the two sides of $5x - 5 = A(x-3) + B(x-2)$ are polynomials that agree at every other $x$⁠, and two polynomials that agree at infinitely many points are identical, so they agree at 2 and 3 as well.)
   Watch the sign of the constant term: $-3A - 2B = -5$ is the same as $3A + 2B = 5$⁠. Lesson: partial fractions

5. Integrate term by term

   By Property V,

   $$\begin{aligned} \int \frac{x^2+1}{x^2-5x+6}\,dx &= \int dx - 5\int \frac{dx}{x-2} \\ &\quad + 10\int \frac{dx}{x-3}. \end{aligned}$$

   For $\int \frac{dx}{x-2}$⁠, substitute $t = x - 2$⁠, $dt = dx$⁠: it becomes $\int \frac{dt}{t} = \log|t|$ by formula (xii), that is $\log|x-2|$⁠. In the same way $\int \frac{dx}{x-3} = \log|x-3|$⁠. Therefore

   $$\begin{aligned} \int \frac{x^2+1}{x^2-5x+6}\,dx &= x - 5\log|x-2| \\ &\quad + 10\log|x-3| + C. \end{aligned}$$
   After the split, every piece is a standard integral: $\int dx = x$⁠, and each $\frac{1}{x-a}$ is a shifted $\frac1x$⁠. The absolute value bars matter: $x - 2$ is negative for $x < 2$⁠, and the log of a negative number is not defined, but $\log|x-2|$ is.
   Standard formula (xii), $\int \frac1x\,dx = \log|x| + C$⁠, with a linear substitution. Basics: logarithms and 1/x

6. Check: differentiate the answer

   Using $\frac{d}{dx}\log|u| = \frac1u$ (for $u \ne 0$⁠):

   $$\begin{aligned} &\frac{d}{dx}\Big[x - 5\log|x-2| + 10\log|x-3|\Big] \\ &= 1 - \frac{5}{x-2} + \frac{10}{x-3}. \end{aligned}$$

   Over the common denominator $(x-2)(x-3)$ the top is

   $$\begin{aligned} &(x-2)(x-3) - 5(x-3) + 10(x-2) \\ &= x^2 - 5x + 6 - 5x + 15 + 10x - 20 \\ &= x^2 + 1. \end{aligned}$$

   So the derivative is $\frac{x^2+1}{x^2-5x+6}$⁠, exactly the integrand.

   An antiderivative is, by definition, a function whose derivative is the integrand. Differentiating back is therefore a complete check, and it catches sign slips in $A$ or $B$ at once.
   $\frac{d}{dx}\log|x| = \frac1x$ for $x \ne 0$ (Section 7.2); the constant $C$ differentiates to 0. Basics: integration undoes differentiation

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