Example 37: $\int\frac{x^4}{(x-1)(x^2+1)}\,dx$

Basics you need

Solution

1. Divide: the fraction is improper

   Expand the denominator: $(x-1)(x^2+1) = x^3 - x^2 + x - 1$⁠. Divide $x^4$ by it:

   • $x^4 \div x^3 = x$⁠; subtract $x(x^3 - x^2 + x - 1) = x^4 - x^3 + x^2 - x$⁠, leaving $x^3 - x^2 + x$⁠.
   • $x^3 \div x^3 = 1$⁠; subtract $1\cdot(x^3 - x^2 + x - 1)$⁠, leaving $1$⁠.

   So the quotient is $x + 1$ and the remainder is 1. Indeed $(x+1)(x^3 - x^2 + x - 1) = x^4 - 1$⁠, so

   $$\frac{x^4}{(x-1)(x^2+1)} = (x + 1) + \frac{1}{(x-1)(x^2+1)} \qquad (1)$$
   Partial fractions only work for proper fractions. Dividing first separates out the polynomial part $x + 1$⁠, which is easy to integrate, and leaves a proper fraction with numerator 1. The quick check $x^4 = (x^4 - 1) + 1$ confirms the division.
   Dividend = divisor × quotient + remainder: $x^4 = (x^3 - x^2 + x - 1)(x + 1) + 1$⁠.

2. Set up the partial fractions

   The factor $x^2 + 1$ cannot be factorised over the reals (its discriminant is $0 - 4 = -4 < 0$⁠). So, by Table 7.2 (form 5),

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

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

   $$\begin{aligned}1 &= A(x^2+1) + (Bx + C)(x - 1)\\ &= (A + B)x^2 + (C - B)x + (A - C) .\end{aligned}$$
   Over a quadratic factor that cannot be split, the numerator must be allowed to be linear, $Bx + C$⁠, not just a constant: a constant alone would not give enough freedom to match every proper fraction.
   Basics: splitting fractions

3. Find $A$⁠, $B$⁠, $C$

   The identity holds for every $x$⁠, so the coefficients on both sides match:

   $$\begin{aligned}x^2:&\quad A + B = 0\\ x:&\quad C - B = 0\\ \text{constant}:&\quad A - C = 1\end{aligned}$$

   From the second, $C = B$⁠; from the first, $B = -A$⁠; so $C = -A$ and the third gives $A + A = 1$⁠. Hence

   $$A = \frac12, \qquad B = C = -\frac12 .$$

   A quicker check for $A$⁠: put $x = 1$ in $1 = A(x^2+1) + (Bx+C)(x-1)$ to get $1 = 2A$⁠. Substituting into (2):

   $$\begin{aligned}&\frac{1}{(x-1)(x^2+1)}\\ &= \frac{1}{2(x-1)} - \frac{x}{2(x^2+1)} - \frac{1}{2(x^2+1)} \quad (3)\end{aligned}$$
   Two polynomials are equal for every $x$ only if their coefficients agree term by term. Three unknowns need three equations, one from each power $x^2$⁠, $x$⁠, $x^0$⁠. We split $\frac{-\frac12 x - \frac12}{x^2+1}$ into two fractions because they integrate differently: one to a logarithm, the other to an inverse tangent.
   Check (3) at $x = 0$⁠: left side $\frac{1}{(-1)(1)} = -1$⁠; right side $-\frac12 - 0 - \frac12 = -1$⁠.

4. Put it all together

   Substituting (3) into (1):

   $$\begin{aligned}&\frac{x^4}{(x-1)(x^2+1)}\\ &= (x+1) + \frac{1}{2(x-1)} - \frac{x}{2(x^2+1)} - \frac{1}{2(x^2+1)} .\end{aligned}$$
   Now every term is one we know how to integrate. (The textbook prints the denominator here and in its final line as $(x-1)(x^2+x+1)$⁠; that is a misprint for $(x-1)(x^2+1)$⁠, the denominator used everywhere else in the solution.)

5. Integrate term by term

   • $\int (x + 1)\,dx = \frac{x^2}{2} + x$ (power rule).
   • $\frac12\int\frac{dx}{x - 1} = \frac12\log|x - 1|$ (standard formula (xii), with $x - 1$ in place of $x$⁠).
   • $\frac12\int\frac{x\,dx}{x^2+1}$⁠: put $u = x^2 + 1$⁠, $du = 2x\,dx$⁠, so it equals $\frac14\int\frac{du}{u} = \frac14\log(x^2+1)$⁠; no modulus is needed since $x^2 + 1 > 0$⁠.
   • $\frac12\int\frac{dx}{x^2+1} = \frac12\tan^{-1}x$ (standard formula (x)).

   Collecting, with the signs from (3):

   $$\begin{aligned}\int \frac{x^4\,dx}{(x-1)(x^2+1)} = {}&\frac{x^2}{2} + x + \frac12\log|x-1|\\ &- \frac14\log(x^2+1) - \frac12\tan^{-1}x + C .\end{aligned}$$
   Integration is linear (Property V, Section 7.2.1), so each term is integrated separately and the constants of integration are gathered into one $C$⁠. The answer is valid on each of the intervals $x < 1$ and $x > 1$⁠.
   Basics: logarithms and 1/x · Basics: substitution

6. Check: $\frac{d}{dx}$ of our answer

   Differentiate term by term:

   $$\begin{aligned}&x + 1 + \frac{1}{2(x-1)} - \frac{2x}{4(x^2+1)} - \frac{1}{2(x^2+1)}\\ &= x + 1 + \frac{1}{2(x-1)} - \frac{x + 1}{2(x^2+1)} .\end{aligned}$$

   Combine the two fractions over $2(x-1)(x^2+1)$⁠:

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

   So the derivative is $(x + 1) + \frac{1}{(x-1)(x^2+1)}$⁠, which by (1) is $\frac{x^4}{(x-1)(x^2+1)}$⁠.

   The derivative of $\log|x - 1|$ is $\frac{1}{x-1}$ on both sides of 1, and that of $\log(x^2+1)$ is $\frac{2x}{x^2+1}$ by the chain rule. Getting back exactly the integrand confirms the division, the partial fractions and the integration all at once.

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