求解积分∫(x²+5)/(x³+3x)dx：常见简便方法及分式分解疑问咨询

Hey there! For this rational function integral, partial fraction decomposition is absolutely the most straightforward and commonly used approach—this is exactly the method you were already working with, and it's the standard go-to for integrals of this type. Let's break down the process clearly:

Step 1: Factor the Denominator #

First, factor the denominator to identify the terms we'll use for partial fractions:

x³ + 3x = x(x² + 3)

Here, x is a linear factor, and x² + 3 is an irreducible quadratic factor (it has no real roots, since its discriminant is negative).

Step 2: Set Up Partial Fractions #

We decompose the rational function into the sum of simpler fractions matching these factors:

(x² + 5)/(x(x² + 3)) = A/x + (Bx + C)/(x² + 3)

Where A, B, and C are constants we need to solve for.

Step 3: Solve for Constants #

Multiply both sides by the denominator x(x² + 3) to eliminate fractions:

x² + 5 = A(x² + 3) + (Bx + C)x

Expand and rearrange the right-hand side:

x² + 5 = (A + B)x² + Cx + 3A

Now equate the coefficients of like terms on both sides to form equations:

• Coefficient of x²: A + B = 1
• Coefficient of x: C = 0
• Constant term: 3A = 5

Solving these gives A = 5/3, B = -2/3, and C = 0.

Step 4: Split and Evaluate the Integral #

Substitute the constants back into the partial fraction decomposition, then split the integral into simpler parts:

∫(x²+5)/(x³+3x)dx = ∫(5/(3x))dx + ∫(-2x/(3(x²+3)))dx

Calculate each integral separately:

1. For the first term:

   ∫5/(3x) dx = (5/3)ln|x| + C₁
   

2. For the second term, use substitution u = x² + 3 (so du = 2x dx):

   ∫(-2x)/(3(x²+3))dx = (-1/3)∫1/u du = (-1/3)ln(x²+3) + C₂
   

   (We can drop the absolute value here since x² + 3 is always positive.)

Step 5: Combine Results #

Add the two integrals together and combine constants into a single C:

(5/3)ln|x| - (1/3)ln(x²+3) + C

You can also use logarithm properties to simplify this further if you prefer:

(1/3)ln(|x|⁵/(x²+3)) + C

Why This Is the Best Method #

Partial fraction decomposition is the standard tool for integrating rational functions because it systematically breaks down complex fractions into elementary terms that we already know how to integrate. For this specific integral, there's no simpler shortcut—substitution alone won't directly simplify the integrand as cleanly as splitting it via partial fractions.

内容的提问来源于stack exchange，提问作者jublikon