Gradshteyn有理函数积分问题求助：相关积分求解指导

Hey there, I totally get how frustrating it is to hammer away at this Gradshteyn rational function integral for days without breaking through—let me walk you through the systematic steps to derive this reduction formula, which is a classic trick for rational integrals with polynomial denominators raised to a power.

This is a standard reduction formula for rational function integrals, and the most reliable way to derive it is either by strategically applying integration by parts, or by assuming the form of the result and solving for unknown coefficients (the latter is more systematic for this case).

Step 1: Define Key Terms & Structure #

First, let's denote the quadratic denominator as:
Q(x) = A + 2Bx + Cx²
Its derivative is straightforward:
Q'(x) = 2B + 2Cx = 2(Cx + B)
Notice this derivative is a linear function, just like the numerator Mx + N—this overlap is the key to building our solution.

Step 2: Assume the Form of the Integral Result #

Since we're dealing with a denominator raised to the power p, we can guess the integral will have two parts:

1. A linear polynomial divided by Q(x)^(p-1) (to account for the derivative of the denominator)
2. A constant multiple of the lower-power integral ∫1/Q(x)^(p-1)dx

Formally, let's write:
$$\int{ (Mx + N) \over Q(x)^{p}}dx={(ax + b) \over Q(x)^{p-1}} + k \int {dx \over Q(x)^{p-1} }$$
Our goal is to solve for coefficients a, b, and k.

Step 3: Differentiate Both Sides & Match Coefficients #

Take the derivative of both sides to eliminate the integral sign:
$$\frac{Mx + N}{Q(x)^p} = \frac{a \cdot Q(x)^{p-1} - (ax + b)(p-1)Q(x)^{{p-2}Q'(x)}{Q(x)}{2(p-1)}} + \frac{k}{Q(x)^{p-1}}$$

Multiply through by Q(x)^p to clear all denominators:
$$Mx + N = aQ(x) - (p-1)(ax + b)Q'(x) + kQ(x)$$

Substitute Q(x) = Cx² + 2Bx + A and Q'(x) = 2(Cx + B) into the right-hand side, then expand and group like terms (x², x, constants). Since the left-hand side has no x² term, we can set up a system of equations by matching coefficients for each term:

1. x² coefficient: (a + k)C - 2(p-1)aC = 0
   Simplifies to: k = a(2p - 3)

2. x coefficient: 2(a + k)B - 2(p-1)(aB + bC) = M

3. Constant term: (a + k)A - 2(p-1)bB = N

Step 4: Solve the Coefficient System #

Using the relation k = a(2p - 3) from the x² term, substitute into the remaining two equations. We end up with a linear system for a and b:
$$\begin{cases}
Ba - Cb = \frac{M}{2(p-1)} \
Aa - Bb = \frac{N}{2(p-1)}
\end{cases}$$

Using Cramer's rule (or substitution) to solve this system, with the determinant D = AC - B² (the discriminant of the quadratic Q(x)):

• $a = \frac{NC - MB}{2(p-1)(AC - B²)}$
• $b = \frac{NB - MA}{2(p-1)(AC - B²)}$
• $k = \frac{(2p-3)(NC - MB)}{2(p-1)(AC - B²)}$

Step 5: Substitute Back to Get the Final Formula #

Plug a, b, and k back into our assumed integral form, and we get exactly the formula you're trying to derive:
$$\int{ (Mx + N) \over(A +2Bx+Cx²⁾{p}}dx={(NB-MA)+(NC-MB)x \over(2(p-1)(AC-B^2)(A+2Bx+Cx2)^{p-1})}+{(2p-3)(NC-MB) \over 2(p-1)(AC-B^2)} \int {dx \over (A+2Bx+Cx²⁾{p-1} }$$

Quick Alternative: Integration by Parts #

If you prefer a more hands-on approach, you can split the numerator Mx + N into a multiple of Q'(x) plus a constant, then integrate each part separately. The first part will be a straightforward substitution, and the second part will require another integration by parts to build the reduction relation for ∫1/Q(x)^p dx—this leads to the same result, but requires a bit more trial-and-error to set up the right u and dv.

内容的提问来源于stack exchange，提问作者Atiqur Rahman