Miscellaneous · Q32: $\int_1^3\frac{dx}{x^2(x+1)}=\frac23+\log\frac23$

Basics you need

Solution

1. The integrand is continuous on $[1, 3]$

   The denominator $x^2(x+1)$ is zero only at $x = 0$ and $x = -1$⁠, both outside $[1, 3]$⁠. On $[1, 3]$ it is at least $1\cdot 2 = 2$⁠. So the integrand is continuous on the closed interval.

   Theorem 2 needs continuity on $[a, b]$⁠. A rational function is continuous wherever its denominator is not zero.
   Lesson: definite integrals and the fundamental theorem

2. Set up the partial fractions

   The numerator 1 has degree 0, less than the degree 3 of the denominator, so the fraction is proper. With the repeated factor $x^2 = (x - 0)^2$ and the simple factor $x + 1$⁠:

   $$\frac{1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} .$$

   Multiply both sides by $x^2(x+1)$⁠:

   $$1 = Ax(x+1) + B(x+1) + Cx^2 .$$
   A repeated factor $x^2$ needs both $\frac{A}{x}$ and $\frac{B}{x^2}$⁠: with only $\frac{B}{x^2}$ there would not be enough freedom to match every such fraction. Clearing the denominators turns the equation into a polynomial identity, true for every $x$ (Remark after Example 11, Section 7.5), so we may substitute any convenient values.
   Table 7.2, form 4. Basics: splitting fractions

3. Find $A$⁠, $B$⁠, $C$

   Put $x = 0$⁠: $1 = B(1)$⁠, so $B = 1$⁠.

   Put $x = -1$⁠: $1 = C(-1)^2$⁠, so $C = 1$⁠.

   Compare the coefficients of $x^2$⁠: the left side has none, the right side has $A + C$⁠. So $A + C = 0$ and $A = -1$⁠.

   $$\frac{1}{x^2(x+1)} = -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1} .$$
   The values $x = 0$ and $x = -1$ each wipe out all but one term, which gives $B$ and $C$ at once. No value of $x$ isolates $A$⁠, so we compare one coefficient instead. Check at $x = 1$⁠: left side $\frac{1}{1\cdot 2} = \frac12$⁠; right side $-1 + 1 + \frac12 = \frac12$ ✓.
   Two polynomials equal for every $x$ have equal coefficients. Lesson: partial fractions

4. Integrate each piece

   $$\begin{aligned} &\int\frac{dx}{x^2(x+1)} \\ &= -\log|x| - \frac{1}{x} + \log|x+1| \\ &=\log\left|\frac{x+1}{x}\right| - \frac1x + C . \end{aligned}$$

   Call this $F(x)$ (without $C$⁠). It is valid for $x \ne 0, -1$⁠, in particular on $[1, 3]$⁠.

   $\int\frac{dx}{x} = \log|x|$ (formula (xii)); $\int x^{-2}dx = \frac{x^{-1}}{-1} = -\frac1x$ (power rule with $n = -2$⁠); $\int\frac{dx}{x+1} = \log|x+1|$ (substitute $u = x + 1$⁠, $du = dx$⁠). The two logs combine by the quotient law.
   Standard formulae (i) and (xii), Section 7.2. Basics: logarithms and 1/x

5. Check: differentiate $F$ back

   $$\begin{aligned} F'(x) &= \frac{1}{x+1} - \frac1x + \frac{1}{x^2} \\ &= \frac{x^2 - x(x+1) + (x+1)}{x^2(x+1)} \\ &= \frac{x^2 - x^2 - x + x + 1}{x^2(x+1)} \\ &= \frac{1}{x^2(x+1)} \ ✓ \end{aligned}$$
   Over the common denominator $x^2(x+1)$ the numerator collapses to 1, which confirms both the partial fractions and the integration. Note $\frac{d}{dx}\left(-\frac1x\right) = +\frac{1}{x^2}$⁠.
   Basics: integration undoes differentiation

6. Evaluate between 1 and 3

   On $[1, 3]$ both $x$ and $x + 1$ are positive, so the bars can be dropped.

   $F(3) = \log\frac43 - \frac13$⁠, and $F(1) = \log 2 - 1$⁠.

   $$\begin{aligned} I &= \left(\log\tfrac43 - \tfrac13\right) - (\log 2 - 1) \\ &= \log\tfrac43 - \log 2 + 1 - \tfrac13 \\ &= \log\tfrac{4}{6} + \tfrac23 = \tfrac23 + \log\tfrac23 . \end{aligned}$$

   This is the required result. As a decimal, $I \approx 0.666667 - 0.405465$ $= 0.261202$⁠.

   Theorem 2: upper value minus lower value. The two logarithms combine by $\log p - \log q = \log\frac{p}{q}$ with $\frac{4/3}{2} = \frac46 = \frac23$⁠. Note $\log\frac23 < 0$⁠, but the total is positive, as it must be for a positive integrand.
   Theorem 2, Section 7.8.3; laws of logarithms. Basics: logarithms

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