Exercise 7.5 · Q13: $\int\frac{2}{(1 - x)(1 + x^2)}\,dx$

Basics you need

Solution

1. The fraction is proper, and $1 + x^2$ cannot be factorised

   The numerator 2 has degree 0 and the denominator $(1 - x)(1 + x^2)$ has degree 3, so the fraction is proper. The quadratic $x^2 + 1$ has discriminant $0^2 - 4(1)(1) = -4 < 0$⁠, so it has no real roots and no real linear factors.

   A proper fraction can be split directly, with no polynomial part. Because $1 + x^2$ cannot be broken into real linear factors, it has to stay as a quadratic factor: this is exactly row 5 of Table 7.2.
   Section 7.5. Lesson: partial fractions

2. Write the form of the decomposition

   $$\frac{2}{(1 - x)(1 + x^2)} = \frac{A}{1 - x} + \frac{Bx + C}{1 + x^2} \qquad (1)$$

   with real constants $A$⁠, $B$⁠, $C$ still to be found.

   The factor $1 - x$ equals $-(x - 1)$⁠, so $\frac{A}{1 - x}$ is just $\frac{-A}{x - 1}$⁠: still a constant over a linear factor, as row 5 requires. Keeping $1 - x$ as printed avoids sign slips. Over the quadratic factor the top must be a general linear expression $Bx + C$⁠, not a single constant.
   Table 7.2, row 5. Basics: splitting fractions

3. Clear the denominators and expand

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

   $$2 = A(1 + x^2) + (Bx + C)(1 - x).$$

   Expand the right side:

   $$\begin{aligned} &A + Ax^2 + Bx - Bx^2 + C - Cx \\ &= (A - B)x^2 + (B - C)x + (A + C). \end{aligned}$$
   (1) is an identity (true for every $x \ne 1$⁠), so after clearing fractions the two polynomials agree at infinitely many points; two polynomials that agree at infinitely many points are the same polynomial, so the cleared equation holds for every real $x$⁠, including $x = 1$⁠. Grouping by powers of $x$ prepares the comparison of coefficients.

4. Find $A$⁠, $B$⁠, $C$

   Compare with the left side $0\cdot x^2 + 0\cdot x + 2$⁠:

   $$\begin{aligned} x^2:&\quad A - B = 0, \\ x:&\quad B - C = 0, \\ \text{constant}:&\quad A + C = 2. \end{aligned}$$

   The first two give $A = B = C$⁠; then $A + A = 2$⁠, so $A = 1$⁠. Hence $A = B = C = 1$⁠.

   Quick cross-check: put $x = 1$ in the cleared identity: $2 = A(1 + 1) + 0$⁠, so $A = 1$⁠. ✓

   Equal polynomials have equal coefficients. Substituting the root $x = 1$ of the linear factor kills the $(Bx + C)(1 - x)$ term and gives $A$ at once; the cleared identity holds for all $x$⁠, including $x = 1$⁠.
   Basics: splitting fractions

5. The integrand as three simple pieces

   $$\frac{2}{(1 - x)(1 + x^2)} = \frac{1}{1 - x} + \frac{x + 1}{1 + x^2},$$

   and splitting the second fraction,

   $$\frac{x + 1}{1 + x^2} = \frac{x}{1 + x^2} + \frac{1}{1 + x^2}.$$
   The piece $\frac{x}{1 + x^2}$ is (half) a derivative-over-function, which gives a logarithm; the piece $\frac{1}{1 + x^2}$ is the inverse tangent form. They need different formulas, so we separate them.

6. Integrate $\frac{1}{1 - x}$

   Substitute $t = 1 - x$⁠, so $dt = -dx$⁠:

   $$\int\frac{dx}{1 - x} = -\int\frac{dt}{t} = -\log|1 - x| + C_1.$$
   The inner function $1 - x$ has derivative $-1$⁠, which produces the minus sign. Forgetting it is the most common slip in this problem.
   Substitution, Section 7.3.1; standard formula (xii), Section 7.2. Basics: substitution

7. Integrate $\frac{x}{1 + x^2}$ and $\frac{1}{1 + x^2}$

   For the first, substitute $t = 1 + x^2$⁠, so $dt = 2x\,dx$⁠, that is $x\,dx = \frac12\,dt$⁠:

   $$\int\frac{x\,dx}{1 + x^2} = \frac12\int\frac{dt}{t} = \frac12\log(1 + x^2) + C_2.$$

   For the second, the standard formula gives

   $$\int\frac{dx}{1 + x^2} = \tan^{-1}x + C_3.$$
   $t = 1 + x^2$ is always positive, so $\log|t| = \log(1 + x^2)$ needs no modulus. The second integral is standard formula (x) of Section 7.2 (or Section 7.4, formula (3) with $a = 1$⁠).
   Section 7.3.1; standard formulae (xii) and (x), Section 7.2. Basics: logarithms and 1/x

8. Put the pieces together

   $$\begin{aligned} &\int\frac{2\,dx}{(1 - x)(1 + x^2)} \\ &= -\log|1 - x| + \tfrac12\log(1 + x^2) \\ &\quad + \tan^{-1}x + C. \end{aligned}$$

   Since $|1 - x| = |x - 1|$⁠, this is the textbook's form $-\log|x - 1| + \frac12\log(1 + x^2) + \tan^{-1}x + C$⁠.

   The three constants $C_1, C_2, C_3$ combine into one arbitrary constant $C$⁠. A number and its negative have the same absolute value, so the two ways of writing the first logarithm are identical.

9. Check: differentiate the answer

   Chain rule on the first term (inner $1 - x$⁠, derivative $-1$⁠) and on the second (inner $1 + x^2$⁠, derivative $2x$⁠):

   $$\begin{aligned} &\frac{d}{dx}\Big(-\log|1 - x| + \tfrac12\log(1 + x^2) + \tan^{-1}x\Big) \\ &= \frac{1}{1 - x} + \frac{x}{1 + x^2} + \frac{1}{1 + x^2} \\ &= \frac{(1 + x^2) + (x + 1)(1 - x)}{(1 - x)(1 + x^2)} \\ &= \frac{1 + x^2 + 1 - x^2}{(1 - x)(1 + x^2)} \\ &= \frac{2}{(1 - x)(1 + x^2)}. \ \checkmark \end{aligned}$$

   Valid for every $x \ne 1$⁠.

   $\frac{d}{dx}\big(-\log|1 - x|\big) = -\frac{1}{1 - x}\cdot(-1) = \frac{1}{1 - x}$⁠. Over the common denominator, $(x + 1)(1 - x) = 1 - x^2$⁠, and the $x^2$ terms cancel, leaving the original numerator 2.
   Property (I), Section 7.2.1. Basics: integration undoes differentiation

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