如何通过换元法求解积分∫1/(x²+x+1)dx？

Hey there! Let’s walk through solving the integral I = ∫1/(x²+x+1)dx using substitution, step by step. The key here is first simplifying the quadratic denominator so we can leverage a standard integral formula after substitution.

First, we need to rewrite the quadratic in the denominator into a form that’s easier to work with via substitution. Completing the square is the perfect tool here:

x² + x + 1 = x² + x + (1/4) + 1 - (1/4)

The first three terms form a perfect square: (x + 1/2)², and the constants simplify to 3/4 (which we can write as (√3/2)²). Now our denominator looks like:

(x + 1/2)² + (√3/2)²

Let’s set u = x + 1/2. Taking the derivative of both sides gives du = dx — ideal, since it directly replaces the dx in our integral. This substitution transforms the denominator into the familiar form u² + a², where a = √3/2.

Substitute u and du into the original integral:

I = ∫1/(u² + (√3/2)²) du

We know the standard result for integrals of the form ∫1/(u² + a²) du is (1/a)arctan(u/a) + C. Plugging in a = √3/2:

I = (1/(√3/2)) arctan(u/(√3/2)) + C

Simplify the coefficients: 1/(√3/2) = 2/√3, and u/(√3/2) = 2u/√3. This simplifies to:

I = (2/√3) arctan(2u/√3) + C

For a cleaner, rationalized form, we can rewrite 2/√3 as 2√3/3:

I = (2√3/3) arctan(2u/√3) + C

Now replace u with our original substitution u = x + 1/2:

I = (2√3/3) arctan( (2(x + 1/2))/√3 ) + C

Simplify the numerator inside the arctangent: 2(x + 1/2) = 2x + 1. So our final result is:

I = (2√3/3) arctan( (2x + 1)/√3 ) + C

Quick sanity check: If you differentiate this result with respect to x, you should end up back with 1/(x²+x+1) — that’s a great way to confirm you didn’t make any arithmetic errors along the way.

内容的提问来源于stack exchange，提问作者user518960