Miscellaneous · Q21: $\int\frac{x^2+x+1}{(x+1)^2(x+2)}\,dx$

Basics you need

Solution

1. Choose the method and the form

   The numerator has degree 2 and the denominator $(x+1)^2(x+2)$ has degree 3, so the fraction is proper. Its factors are $x+1$ (twice) and $x+2$⁠, so

   $$\frac{x^2+x+1}{(x+1)^2(x+2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+2} . \qquad (1)$$
   A rational function with a factorised denominator and no obvious substitution is a job for partial fractions. A repeated factor $(x+1)^2$ needs two terms, one over $x+1$ and one over $(x+1)^2$⁠; with only the squared term there would be just two unknowns, too few to match the three coefficients of a general quadratic numerator.
   Section 7.5, Table 7.2, row 4. Lesson: partial fractions

2. Clear the denominators

   Multiply (1) by $(x+1)^2(x+2)$⁠:

   $$x^2 + x + 1 = A(x+1)(x+2) + B(x+2) + C(x+1)^2 . \qquad (2)$$
   (1) holds for every $x \ne -1, -2$⁠, so the two polynomials in (2) agree at infinitely many points and therefore for every real $x$⁠. That is what lets us put in $x = -1$ and $x = -2$ next.
   Remark after Example 11, Section 7.5.

3. Find $A$⁠, $B$⁠, $C$

   $x = -1$⁠: $1 - 1 + 1 = B(1)$⁠, so $B = 1$⁠.

   $x = -2$⁠: $4 - 2 + 1 = C(-1)^2$⁠, so $C = 3$⁠.

   Coefficient of $x^2$ in (2): $1 = A + C$⁠, so $A = 1 - 3 = -2$⁠.

   Confirm with the constant term: the right side of (2) has constant $2A + 2B + C = -4 + 2 + 3 = 1$⁠, matching the left side.

   Each of the two roots kills all but one term. No value of $x$ isolates $A$⁠, so we compare the $x^2$ coefficients (only $A(x+1)(x+2)$ and $C(x+1)^2$ contain $x^2$⁠). The constant-term comparison is an independent check of all three values.
   Basics: splitting fractions

4. Integrate each piece

   $$\frac{x^2+x+1}{(x+1)^2(x+2)} = -\frac{2}{x+1} + \frac{1}{(x+1)^2} + \frac{3}{x+2} ,$$ $$\begin{aligned}&\int\frac{x^2+x+1}{(x+1)^2(x+2)}\,dx\\ &= -2\log|x+1| - \frac{1}{x+1} + 3\log|x+2| + C .\end{aligned}$$
   $\int\frac{dx}{x+1} = \log|x+1|$ and $\int\frac{dx}{x+2} = \log|x+2|$ (substitute $t = x+k$⁠, standard formula (xii)). For the middle term, $\int(x+1)^{-2}\,dx = \frac{(x+1)^{-1}}{-1} = -\frac{1}{x+1}$ (power formula with $n = -2$⁠): not a logarithm.
   Standard formulae (i) and (xii), Section 7.2. Basics: logarithms and 1/x

5. Check: differentiate the answer

   $$\begin{aligned}&\frac{d}{dx}\left(-2\log|x+1| - \frac{1}{x+1} + 3\log|x+2|\right)\\ &= -\frac{2}{x+1} + \frac{1}{(x+1)^2} + \frac{3}{x+2} .\end{aligned}$$

   Over the common denominator $(x+1)^2(x+2)$ the numerator is

   $$\begin{aligned}&-2(x+1)(x+2) + (x+2) + 3(x+1)^2\\ &= (-2x^2 - 6x - 4) + (x + 2) + (3x^2 + 6x + 3)\\ &= x^2 + x + 1 .\ \checkmark\end{aligned}$$
   The derivative is the integrand. The answer is valid on each of the intervals $(-\infty, -2)$⁠, $(-2, -1)$ and $(-1, \infty)$⁠, which avoid both excluded points.
   $\frac{d}{dx}\log|x| = \frac1x$ (Section 7.2, list item (xii)). Basics: integration undoes differentiation

6. Compare with the textbook's answer

   The key gives $-2\log|x+1| - \frac{1}{x+1} + 3\log|x+2| + C$⁠, exactly our answer.

   Partial fractions are unique, so any correct method must give these same three terms; only the constant can differ.

Every fact used, and where it comes from