Integration by partial fractions
No rule integrates five x plus one over x squared plus x minus two as it stands. How do you take a fraction like that apart into pieces you already know how to integrate?
Where you are 16 in Integrals
When you reach for it
Adding fractions runs one way: \dfrac{3}{x + 2} + \dfrac{2}{x - 1} combines over the common denominator into \dfrac{5x + 1}{x^2 + x - 2}. Integration needs the reverse. The combined fraction matches no rule, but each piece is a constant over a linear factor, whose antiderivative is a logarithm. Partial fractions is the method that recovers the pieces from the combined fraction.
Reach for it when the integrand is a rational function (a polynomial over a polynomial) whose denominator factors, and no simpler move such as u-substitution fits. Day 1 covers denominators that split into distinct linear factors. Day 2 adds repeated linear factors and quadratic factors with no real roots.
The procedure
Day 1: distinct linear factors
Proper first. A rational function \dfrac{P(x)}{Q(x)} is proper when the degree of P is less than the degree of Q. If it is not, divide first: the quotient is a polynomial, integrated term by term, and only the proper remainder is split.
The decomposition. Suppose Q(x) = c(x - r_1)(x - r_2)\cdots(x - r_n) with the r_i distinct real numbers, and P has degree less than n. Then there are unique constants A_1, \dots, A_n with
A factor ax + b is \displaystyle a\left(x + \frac{b}{a}\right), so the statement covers it too, with \dfrac{A}{ax + b} as its piece.
Why the constants exist. Multiply through by Q(x). The equation becomes the polynomial identity
Put x = r_i: every term on the right but the i-th contains the factor x - r_i and vanishes, so the identity forces
That proves uniqueness. For existence, choose the A_i this way and call the right side R(x). Then R has degree less than n and agrees with P at the n points r_1, \dots, r_n, so P - R is a polynomial of degree less than n with n roots. By the The remainder and factor theorems it would have degree at least n unless it is the zero polynomial, so P = R for every x.
One point needs care. The original equation holds only where Q(x) \ne 0, yet the argument substitutes the roots. That is allowed because the cleared equation is between two polynomials that agree at infinitely many points, and two such polynomials have identical coefficients, so they agree everywhere.
Two ways to find the constants. The formula for A_i is the cover-up method: to find the constant over x - r_i, cover that factor in \dfrac{P(x)}{Q(x)} and evaluate what is left at x = r_i. The other way is matching coefficients: expand the cleared identity and set the coefficient of each power of x equal on both sides, which gives a linear system. Cover-up is faster for distinct linear factors; matching coefficients always works and is what Day 2 needs.
Integrating the pieces. By u-substitution with u = ax + b,
The answer holds on each interval between roots of Q, where the integrand is continuous.
Day 2: repeated and quadratic factors
A repeated linear factor. A factor (x - r)^m in Q contributes one piece for every power up to m:
The top power alone is not enough. For example \dfrac{x}{(x - 1)^2} = \dfrac{1}{x - 1} + \dfrac{1}{(x - 1)^2}, since x = (x - 1) + 1, and no single constant over (x - 1)^2 equals it.
An irreducible quadratic factor. A quadratic x^2 + px + q with p^2 - 4q < 0 has no real roots and does not factor over the real numbers. Its piece has a linear numerator, \dfrac{Bx + C}{x^2 + px + q}, because anything of degree less than 2 may sit over it. A repeated factor \left(x^2 + px + q\right)^m contributes \dfrac{B_k x + C_k}{\left(x^2 + px + q\right)^k} for k = 1, \dots, m.
The general statement. Every real polynomial factors into real linear factors and irreducible quadratics; that follows from the The fundamental theorem of algebra and the fact that non-real roots come in conjugate pairs. A proper \dfrac{P(x)}{Q(x)} then equals the sum of the pieces above, one set for each factor, with unique real constants. The number of unknown constants always equals the degree of Q. This is stated here without proof; the proof is a theorem of algebra about the resulting linear system. Cover-up still finds the constant over the highest power of each linear factor, and the remaining constants come from matching coefficients or from substituting other convenient values of x.
Integrating the pieces.
| Piece | Antiderivative |
|---|---|
| \dfrac{A}{(x - r)^k}, k \ge 2 | -\dfrac{A}{(k - 1)(x - r)^{k - 1}} |
| \dfrac{Bx + C}{x^2 + px + q} | a log part plus an arctangent part |
| \dfrac{Bx + C}{\left(x^2 + px + q\right)^k}, k \ge 2 | see below |
The first row is the power rule. The second is the move in Completing the square inside an integral: split Bx + C into a multiple of 2x + p plus a constant, so the first part gives \ln\left(x^2 + px + q\right) and the second gives an arctangent. In the third row the multiple of 2x + p is again a u-substitution, giving a power of the quadratic; a leftover constant over \left(x^2 + px + q\right)^k needs Trigonometric substitution. So every rational function has an antiderivative built from rational functions, logarithms and arctangents.
Worked examples
Day 1: two factors, by cover-up
Find \int \dfrac{5x + 1}{x^2 + x - 2}\,dx.
SolutionThe integrand is proper, and x^2 + x - 2 = (x + 2)(x - 1), two distinct linear factors. So
Cover-up: cover x + 2 and put x = -2 in what is left, \dfrac{5x + 1}{x - 1}, which gives A = \dfrac{-9}{-3} = 3. Cover x - 1 and put x = 1 in \dfrac{5x + 1}{x + 2}, which gives B = \dfrac{6}{3} = 2. Check by recombining: 3(x - 1) + 2(x + 2) = 5x + 1. Then
on each of the intervals x < -2, -2 < x < 1 and x > 1.
Day 1: three factors, by matching coefficients
Find \int \dfrac{10x^2 - 1}{2x^3 + x^2 - x}\,dx.
SolutionFactor out x, then factor the quadratic: 2x^3 + x^2 - x = x\left(2x^2 + x - 1\right) = x(2x - 1)(x + 1). The numerator has degree 2 < 3, so
Clearing denominators and expanding each product:
Matching the coefficients of x^2, x and 1:
So A = 1, then B - C = -1 and B + 2C = 8; subtracting gives 3C = 9, so C = 3 and B = 2. Cover-up agrees: at \displaystyle x = \frac{1}{2}, covering 2x - 1 leaves \dfrac{10x^2 - 1}{x(x + 1)} = \dfrac{3/2}{3/4} = 2. The middle piece has a = 2, so it brings a factor \displaystyle \frac{1}{2}:
Day 1: dividing first
Find \int \dfrac{x^3}{x^2 - 1}\,dx.
SolutionThe numerator’s degree is not less than the denominator’s, so divide: x^3 = x\left(x^2 - 1\right) + x, which gives
The remainder is proper, and x^2 - 1 = (x - 1)(x + 1). Cover-up gives \dfrac{1}{2} over x - 1 (put x = 1 in \dfrac{x}{x + 1}) and \dfrac{-1}{-2} = \dfrac{1}{2} over x + 1. Therefore
The two logs combine into \displaystyle \frac{1}{2}\ln\left\lvert x^2 - 1 \right\rvert, which u-substitution with u = x^2 - 1 also gives directly, since the remainder’s numerator is half the derivative of its denominator. Partial fractions does not need that coincidence.
Day 1: Lalaina’s reaction
In a chemistry lab Lalaina mixes substance A at 2 moles per liter with substance B at 5 moles per liter. Each reaction uses one unit of each, so once x moles per liter of product have formed, 2 - x of A and 5 - x of B remain, and the rate is
moles per liter per minute. How long does it take to form 1 mole per liter of product?
SolutionWhile x < 2 the rate is positive, so x increases with t and t can be read as a function of x. By the Derivatives of inverse functions, \dfrac{dt}{dx} = \dfrac{1}{dx/dt} = \dfrac{20}{(2 - x)(5 - x)}, and the time from x = 0 to x = 1 is the definite integral of \dfrac{dt}{dx}:
Cover-up: at x = 2, covering 2 - x leaves \dfrac{1}{5 - x} = \dfrac{1}{3}; at x = 5, covering 5 - x leaves \dfrac{1}{2 - x} = -\dfrac{1}{3}. So
Here a = -1 in each factor, so \int \dfrac{dx}{2 - x} = -\ln\lvert 2 - x \rvert, and an antiderivative is \dfrac{1}{3}\ln\left\lvert \dfrac{5 - x}{2 - x} \right\rvert. On [0, 1] both factors are positive, and
The same integral up to X gives \displaystyle t(X) = \frac{20}{3}\ln\dfrac{2(5 - X)}{5(2 - X)}, which grows without bound as X \to 2^-. The product never quite reaches 2 moles per liter, the amount at which A runs out (Figure 1).
Day 2: a repeated linear factor
Find \int \dfrac{x^2 + 2}{x(x - 1)^2}\,dx.
SolutionThe factor x - 1 appears twice, so it contributes two pieces:
Clearing denominators gives x^2 + 2 = A(x - 1)^2 + Bx(x - 1) + Cx. At x = 0: 2 = A. At x = 1: 3 = C. No value of x isolates B, so match the coefficient of x^2: 1 = A + B, so B = -1. The piece \dfrac{3}{(x - 1)^2} integrates by the power rule to -\dfrac{3}{x - 1}:
Check: the derivative is \dfrac{2}{x} - \dfrac{1}{x - 1} + \dfrac{3}{(x - 1)^2}, whose numerator over x(x - 1)^2 is 2(x - 1)^2 - x(x - 1) + 3x = x^2 + 2.
Day 2: an irreducible quadratic factor
Find \int \dfrac{3x^2 - x + 4}{x^3 + 4x}\,dx.
Solutionx^3 + 4x = x\left(x^2 + 4\right), and x^2 + 4 has no real roots. So
Clearing denominators gives 3x^2 - x + 4 = A\left(x^2 + 4\right) + (Bx + C)x. At x = 0: 4 = 4A, so A = 1. Matching x^2: 3 = A + B, so B = 2. Matching x: C = -1. Then
The middle piece is the derivative of x^2 + 4 over itself, and the last is an arctangent form, since \int \dfrac{dx}{x^2 + 4} = \dfrac{1}{2}\arctan\dfrac{x}{2} + C:
No absolute value is needed on x^2 + 4, which is always positive.
Day 2: a repeated quadratic factor, over an interval
Evaluate \int_0^1 \dfrac{2x^3 + x^2 + 4x + 1}{\left(x^2 + 1\right)^2}\,dx.
SolutionThe factor x^2 + 1 is irreducible and appears twice, so
There is no real root to substitute, so match coefficients in
So A = 2, B = 1, A + C = 4 gives C = 2, and B + D = 1 gives D = 0. The pieces are \dfrac{2x}{x^2 + 1}, \dfrac{1}{x^2 + 1} and \dfrac{2x}{\left(x^2 + 1\right)^2}. With u = x^2 + 1 the last is \int u^{-2}\,du = -u^{-1}, so an antiderivative is
The integrand is continuous on [0, 1], so
Since D = 0, no trigonometric substitution was needed here.
Boundaries
- The fraction must be proper. The decomposition theorem needs \deg P < \deg Q. Applied to an improper fraction, cover-up still returns numbers, but their sum is a different function (Exercise 10). Divide first.
- The denominator must be factored, and that can be the hard part. The theorem says real factors exist, not that they can be found exactly. A cubic such as x^3 + x + 1 has one real root that is not rational; the cubic formula gives it only as a tangle of nested cube roots, and in practice it is found numerically.
- Irreducible means no real roots. A quadratic is irreducible only when its discriminant is negative. \displaystyle x^2 - 2 = \left(x - \sqrt{2}\right)\left(x + \sqrt{2}\right) splits into linear factors, and treating it as a quadratic piece leaves the decomposition incomplete.
- Each antiderivative holds between roots of Q. A term \ln\lvert x - r \rvert or \dfrac{1}{x - r} breaks at x = r, and the constant C may differ from one interval to the next. A definite integral by the fundamental theorem needs [a, b] to contain no root of Q; otherwise it is an improper integral (Exercise 11).
- A constant over a repeated quadratic needs another method. \int \dfrac{dx}{\left(x^2 + 1\right)^2} is not reached by logs and arctangents directly; it needs Trigonometric substitution.
Where this goes next
Logistic growth leads to an integral of \dfrac{1}{P(K - P)}, which splits exactly as Lalaina’s reaction did, into two pieces over linear factors. Series use the same split on terms such as \dfrac{1}{n(n + 1)} = \dfrac{1}{n} - \dfrac{1}{n + 1}, where most of the pieces cancel in the running total. Trigonometric substitution finishes the one case left open here, a constant over a power of an irreducible quadratic. Choosing an integration technique places partial fractions among the other methods: it is the method for a rational integrand once simpler substitutions have been ruled out.
Exercises
Day 1: distinct linear factors
- Find \int \dfrac{dx}{x^2 - 16}.
- Find \int \dfrac{3x - 1}{x^2 + 2x - 15}\,dx.
- Find \int \dfrac{x - 9}{x^2 - 3x}\,dx.
- Find \int \dfrac{5}{2x^2 + 7x + 3}\,dx.
- Find \int \dfrac{2x^2 - x + 2}{x^3 + x^2 - 2x}\,dx.
- Find \int \dfrac{x^2 + 1}{x^2 - x - 6}\,dx.
- Evaluate \int_4^6 \dfrac{dx}{x^2 - 4x + 3}.
- In Lalaina’s reaction (Example 4), how long does it take to form 1.5 moles per liter of product? How long does the half mole from x = 1 to x = 1.5 take, compared with the first full mole?
Day 1: reasoning
- For r \ne s, show that \dfrac{1}{(x - r)(x - s)} = \dfrac{1}{r - s}\left(\dfrac{1}{x - r} - \dfrac{1}{x - s}\right), and find \int \dfrac{dx}{(x - r)(x - s)}. What happens to the formula when r = s, and what is the integral then?
- Apply cover-up to \dfrac{x^2}{(x - 1)(x + 1)} as if it were proper. Show that the resulting sum is not equal to the fraction, then find the correct decomposition and \int \dfrac{x^2}{x^2 - 1}\,dx.
- Andry evaluates \int_0^3 \dfrac{dx}{(x - 1)(x - 2)} as \ln\left\lvert \dfrac{x - 2}{x - 1} \right\rvert \Big|_0^3 = -2\ln 2. Check that his function is an antiderivative, and explain why the answer is still not valid.
Day 2: repeated and quadratic factors
- Find \int \dfrac{x + 5}{(x + 1)^2}\,dx.
- Find \int \dfrac{3x^2 - 2x + 1}{x^3 - x^2}\,dx.
- Find \int \dfrac{dx}{x(x + 2)^2}.
- Find \int \dfrac{3x^2 - 2x + 3}{x^3 - x^2 + x - 1}\,dx.
- Find \int \dfrac{5x^2 + 8x + 11}{(x + 1)\left(x^2 + 2x + 5\right)}\,dx.
- Find \int \dfrac{x^3 + x^2 + 4x + 2}{\left(x^2 + 2\right)^2}\,dx.
- Evaluate \int_1^2 \dfrac{dx}{x^2(x + 1)}.
- Mirindra runs the reaction with both substances at 3 moles per liter, so the rate is \dfrac{dx}{dt} = 0.1(3 - x)^2 moles per liter per minute. How long does it take to form 2 moles per liter of product?
Day 2: reasoning
- Write the form of each decomposition without finding the constants, and count the unknowns: (a) \dfrac{x^2 + 1}{(x - 2)^3\left(x^2 + x + 1\right)}; (b) \dfrac{x^3 + 1}{x^2\left(x^2 + 9\right)^2}; (c) \dfrac{1}{x\left(x^2 - 2\right)}; (d) \dfrac{x^4 + 1}{x^3 + x}.
- Mirindra tries \dfrac{x + 3}{x(x + 1)^2} = \dfrac{A}{x} + \dfrac{C}{(x + 1)^2}. Show that no constants A and C work, then find the correct decomposition.
- In Exercise 15, the cleared identity is 3x^2 - 2x + 3 = A\left(x^2 + 1\right) + (Bx + C)(x - 1). Substitute x = i to find B and C at once, and explain why one complex value gives two equations.
Answers
- \dfrac{1}{8}\ln\lvert x - 4 \rvert - \dfrac{1}{8}\ln\lvert x + 4 \rvert + C. Cover-up gives \displaystyle \frac{1}{8} over x - 4 and \displaystyle -\frac{1}{8} over x + 4.
- 2\ln\lvert x + 5 \rvert + \ln\lvert x - 3 \rvert + C. Here x^2 + 2x - 15 = (x + 5)(x - 3); at x = -5, \displaystyle \frac{-16}{-8} = 2; at x = 3, \displaystyle \frac{8}{8} = 1.
- 3\ln\lvert x \rvert - 2\ln\lvert x - 3 \rvert + C. At x = 0, \displaystyle \frac{-9}{-3} = 3; at x = 3, \displaystyle \frac{-6}{3} = -2.
- \ln\lvert 2x + 1 \rvert - \ln\lvert x + 3 \rvert + C. The denominator is (2x + 1)(x + 3), and \displaystyle \frac{5}{(2x + 1)(x + 3)} = \frac{2}{2x + 1} - \frac{1}{x + 3}; the first piece integrates to \displaystyle \frac{2}{2}\ln\lvert 2x + 1 \rvert.
- -\ln\lvert x \rvert + \ln\lvert x - 1 \rvert + 2\ln\lvert x + 2 \rvert + C. The denominator is x(x - 1)(x + 2); cover-up gives \displaystyle \frac{2}{-2} = -1, \displaystyle \frac{3}{3} = 1 and \displaystyle \frac{12}{6} = 2.
- x + 2\ln\lvert x - 3 \rvert - \ln\lvert x + 2 \rvert + C. Division gives \displaystyle 1 + \frac{x + 7}{(x - 3)(x + 2)}, and \displaystyle \frac{x + 7}{(x - 3)(x + 2)} = \frac{2}{x - 3} - \frac{1}{x + 2}.
- \dfrac{1}{2}\ln\dfrac{9}{5}. The antiderivative \displaystyle \frac{1}{2}\ln\left\lvert\frac{x - 3}{x - 1}\right\rvert is valid on [4, 6], which avoids 1 and 3; it gives \displaystyle \frac{1}{2}\ln\frac{3}{5} - \frac{1}{2}\ln\frac{1}{3}.
- \dfrac{20}{3}\ln\dfrac{14}{5} \approx 6.86 minutes in all; the half mole from 1 to 1.5 takes \dfrac{20}{3}\ln\dfrac{7}{4} \approx 3.73 minutes, longer than the 3.13 minutes for the whole first mole. At x = 1.5, \displaystyle \frac{5 - x}{2 - x} = 7, and \displaystyle \frac{20}{3}\left(\ln 7 - \ln\frac{5}{2}\right) = \frac{20}{3}\ln\frac{14}{5}; subtracting \displaystyle \frac{20}{3}\ln\frac{8}{5} leaves \displaystyle \frac{20}{3}\ln\frac{7}{4}. The reaction slows as A runs out.
- Combining the right side over (x - r)(x - s) gives numerator \displaystyle \frac{(x - s) - (x - r)}{r - s} = 1. So \int \dfrac{dx}{(x - r)(x - s)} = \dfrac{1}{r - s}\ln\left\lvert \dfrac{x - r}{x - s} \right\rvert + C. When r = s the factor \displaystyle \frac{1}{r - s} is undefined: the factors are no longer distinct, the integrand is \displaystyle \frac{1}{(x - r)^2}, and the integral is -\dfrac{1}{x - r} + C.
- Cover-up gives \displaystyle \frac{1}{2} over x - 1 and \displaystyle -\frac{1}{2} over x + 1, whose sum is \dfrac{1}{x^2 - 1}, not \dfrac{x^2}{x^2 - 1} (at x = 0 the first is -1 and the second is 0). Dividing first, \dfrac{x^2}{x^2 - 1} = 1 + \dfrac{1}{x^2 - 1}, so the integral is x + \dfrac{1}{2}\ln\lvert x - 1 \rvert - \dfrac{1}{2}\ln\lvert x + 1 \rvert + C.
- The derivative of \ln\lvert x - 2 \rvert - \ln\lvert x - 1 \rvert is \displaystyle \frac{1}{x - 2} - \frac{1}{x - 1} = \frac{1}{(x - 1)(x - 2)}, so it is an antiderivative on each interval avoiding 1 and 2. But [0, 3] contains both, where the integrand is unbounded, so the fundamental theorem does not apply. The integral is improper, and near x = 1 the integrand behaves like \displaystyle -\frac{1}{x - 1}, whose integral diverges; there is no number to find.
- \ln\lvert x + 1 \rvert - \dfrac{4}{x + 1} + C. Since x + 5 = (x + 1) + 4, the fraction is \displaystyle \frac{1}{x + 1} + \frac{4}{(x + 1)^2}.
- \ln\lvert x \rvert + \dfrac{1}{x} + 2\ln\lvert x - 1 \rvert + C. The denominator is x^2(x - 1), and 3x^2 - 2x + 1 = Ax(x - 1) + B(x - 1) + Cx^2 gives B = -1 at x = 0, C = 2 at x = 1, and A = 3 - C = 1 from x^2.
- \dfrac{1}{4}\ln\lvert x \rvert - \dfrac{1}{4}\ln\lvert x + 2 \rvert + \dfrac{1}{2(x + 2)} + C. From 1 = A(x + 2)^2 + Bx(x + 2) + Cx: \displaystyle A = \frac{1}{4} at x = 0, \displaystyle C = -\frac{1}{2} at x = -2, and \displaystyle B = -A = -\frac{1}{4} from x^2.
- 2\ln\lvert x - 1 \rvert + \dfrac{1}{2}\ln\left(x^2 + 1\right) - \arctan x + C. The denominator is x^2(x - 1) + (x - 1) = (x - 1)\left(x^2 + 1\right), and the fraction is \displaystyle \frac{2}{x - 1} + \frac{x - 1}{x^2 + 1}.
- 2\ln\lvert x + 1 \rvert + \dfrac{3}{2}\ln\left(x^2 + 2x + 5\right) - \arctan\dfrac{x + 1}{2} + C. Cover-up at x = -1 gives \displaystyle \frac{8}{4} = 2; then 5x^2 + 8x + 11 - 2\left(x^2 + 2x + 5\right) = 3x^2 + 4x + 1 = (3x + 1)(x + 1), so the quadratic piece is \displaystyle \frac{3x + 1}{x^2 + 2x + 5}. Split \displaystyle 3x + 1 = \frac{3}{2}(2x + 2) - 2, and x^2 + 2x + 5 = (x + 1)^2 + 4.
- \dfrac{1}{2}\ln\left(x^2 + 2\right) \displaystyle + \dfrac{1}{\sqrt{2}}\arctan\dfrac{x}{\sqrt{2}} - \dfrac{1}{x^2 + 2} + C. Matching coefficients gives \displaystyle \frac{x + 1}{x^2 + 2} + \frac{2x}{\left(x^2 + 2\right)^2}.
- \ln\dfrac{3}{4} + \dfrac{1}{2} \approx 0.212. The fraction is \displaystyle -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x + 1}, so \displaystyle F(x) = \ln\frac{x + 1}{x} - \frac{1}{x}, and \displaystyle F(2) - F(1) = \left(\ln\frac{3}{2} - \frac{1}{2}\right) - (\ln 2 - 1).
- \dfrac{20}{3} \approx 6.67 minutes. Here \displaystyle \frac{dt}{dx} = \frac{10}{(3 - x)^2}, a single repeated factor, and \displaystyle \int_0^2 \frac{10}{(3 - x)^2}\,dx = \left[\frac{10}{3 - x}\right]_0^2 = 10 - \frac{10}{3}.
- (a) \displaystyle \frac{A}{x - 2} + \frac{B}{(x - 2)^2} + \frac{C}{(x - 2)^3} \displaystyle + \frac{Dx + E}{x^2 + x + 1}: five unknowns, degree 5 (x^2 + x + 1 has discriminant -3). (b) \displaystyle \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 9} \displaystyle + \frac{Ex + F}{\left(x^2 + 9\right)^2}: six unknowns, degree 6. (c) x^2 - 2 is not irreducible, so the form is \displaystyle \frac{A}{x} + \frac{B}{x - \sqrt{2}} + \frac{C}{x + \sqrt{2}}: three unknowns. (d) Improper, so divide first: x^4 + 1 = x\left(x^3 + x\right) + \left(1 - x^2\right), so the fraction is x plus \displaystyle \frac{1 - x^2}{x\left(x^2 + 1\right)}, and that remainder takes the form \displaystyle \frac{A}{x} + \frac{Bx + C}{x^2 + 1}: three unknowns.
- Clearing gives x + 3 = A(x + 1)^2 + Cx. At x = 0, A = 3; but the coefficient of x^2 on the left is 0, which forces A = 0. No constants work, because the piece \displaystyle \frac{B}{x + 1} is missing. The correct decomposition is \dfrac{3}{x} - \dfrac{3}{x + 1} - \dfrac{2}{(x + 1)^2}: A = 3 at x = 0, C = -2 at x = -1, and B = -A = -3 from x^2.
- B = 1, C = -1. At x = i the term A\left(x^2 + 1\right) vanishes and the left side is -3 - 2i + 3 = -2i, while (Bi + C)(i - 1) = (-B - C) + (C - B)i. Since B and C are real, the real parts and the imaginary parts must match separately: -B - C = 0 and C - B = -2. The identity is between polynomials, so it holds at complex x too.