Exercise 7.5 · Q7: $\int\frac{x}{(x^2+1)(x-1)}\,dx$

Basics you need

Solution

1. Proper, with one linear and one irreducible quadratic factor

   The numerator $x$ has degree 1; the denominator $(x^2+1)(x-1)$ has degree 3. So the fraction is proper. The factor $x - 1$ is linear. The factor $x^2 + 1$ has discriminant $0^2 - 4(1)(1) = -4 < 0$⁠, so it has no real roots and cannot be factorised further.

   Row 5 of Table 7.2 is exactly this case: a proper fraction over (linear) $\times$ (quadratic that cannot be factorised).
   Section 7.5; Table 7.2, row 5. Basics: splitting fractions

2. Write the form of the decomposition

   $$\frac{x}{(x^2+1)(x-1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1} \qquad (1)$$
   Over a quadratic factor the numerator must be allowed to be linear, $Bx + C$⁠. A constant alone would not be enough: we would have only two unknowns for three coefficients, and in general no solution.
   Table 7.2, row 5. Lesson: partial fractions

3. Clear the denominators

   Multiply (1) by $(x^2+1)(x-1)$⁠:

   $$x = A(x^2+1) + (Bx+C)(x-1). \qquad (2)$$
   (1) holds for every $x \ne 1$⁠; the polynomials in (2) agree at infinitely many points, so (2) holds for every real $x$⁠.
   Remark after Example 11, Section 7.5.

4. Find $A$ by putting $x = 1$

   $$1 = A(1 + 1) + (B + C)(0) = 2A, \quad\text{so } A = \tfrac12.$$
   At $x = 1$ the second term contains the factor $x - 1 = 0$ and vanishes. The quadratic $x^2 + 1$ has no real root, so there is no similar shortcut for $B$ and $C$⁠; we compare coefficients instead.

5. Find $B$ and $C$ by comparing coefficients

   Expand the right side of (2):

   $$\begin{aligned} &A x^2 + A + Bx^2 - Bx + Cx - C \\ &= (A+B)x^2 + (C-B)x + (A-C). \end{aligned}$$

   Compare with the left side $0x^2 + 1x + 0$⁠:

   $$\begin{aligned} x^2:&\quad A + B = 0 \;\Rightarrow\; B = -\tfrac12, \\ \text{const}:&\quad A - C = 0 \;\Rightarrow\; C = \tfrac12, \\ x:&\quad C - B = \tfrac12 + \tfrac12 = 1. \ \checkmark \end{aligned}$$
   The $x^2$ and constant equations give $B$ and $C$ at once from $A = \frac12$⁠. The $x$ equation is then a check, and it holds.
   Equal polynomials have equal coefficients. Basics: splitting fractions

6. Write the integrand in pieces

   $$\begin{aligned} \frac{x}{(x^2+1)(x-1)} &= \frac{1}{2(x-1)} + \frac{-\frac12 x + \frac12}{x^2+1} \\ &= \frac{1}{2(x-1)} - \frac12\cdot\frac{x}{x^2+1} \\ &\quad + \frac12\cdot\frac{1}{x^2+1}. \end{aligned}$$
   Separating $\frac{Bx+C}{x^2+1}$ into an $x$ part and a constant part gives two pieces that each have a known integral.

7. Integrate each piece

   For the middle piece put $t = x^2 + 1$⁠, $dt = 2x\,dx$⁠:

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

   (No absolute value is needed since $x^2 + 1 > 0$⁠.) Then

   $$\begin{aligned} \int f(x)\,dx &= \frac12\log|x-1| - \frac12\cdot\frac12\log(x^2+1) \\ &\quad + \frac12\tan^{-1}x + C \\ &= \frac12\log|x-1| - \frac14\log(x^2+1) \\ &\quad + \frac12\tan^{-1}x + C. \end{aligned}$$
   The first piece is $\frac12\int\frac{dx}{x-1}$⁠; the last is $\frac12\int\frac{dx}{1+x^2}$⁠. Constants move outside (Property (IV), Section 7.2.1).
   Substitution, Section 7.3.1; standard formulae (x) and (xii), Section 7.2. Basics: substitution

8. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\Big(\tfrac12\log|x-1| - \tfrac14\log(x^2+1) \\ &\qquad + \tfrac12\tan^{-1}x\Big) \\ &= \frac{1}{2(x-1)} - \frac{2x}{4(x^2+1)} + \frac{1}{2(1+x^2)} \\ &= \frac{1}{2(x-1)} + \frac{1-x}{2(x^2+1)}. \end{aligned}$$

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

   $$\begin{aligned} &(x^2+1) + (1-x)(x-1) \\ &= x^2 + 1 - (x^2 - 2x + 1) = 2x, \end{aligned}$$

   so the derivative is $\frac{2x}{2(x-1)(x^2+1)} = \frac{x}{(x^2+1)(x-1)}$⁠. ✓

   $(1-x)(x-1) = -(x-1)^2$⁠. We are back at the integrand; the answer is valid on each interval avoiding $x = 1$⁠.
   Chain rule; $\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$⁠, Section 7.2, list item (x). Basics: integration undoes differentiation

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