关于$(2x-6)^2$两种积分方法结果不一致的疑问

Hey there! Great question—it’s totally common to get confused when two integration methods seem to give different answers at first glance. Let’s break this down to see why both are actually correct.

First, let’s expand your second result from the reverse chain rule to compare it directly with your first answer:

• Start with $\frac{1}{6}(2x-6)^3$. Let’s factor out the 2 from the parentheses to simplify expansion:
  $$\frac{1}{6}(2(x-3))^3 = \frac{1}{6} \times 8(x-3)^3 = \frac{4}{3}(x^3 - 9x^2 + 27x - 27)$$
• Multiplying through gives:
  $$\frac{4}{3}x^3 - 12x^2 + 36x - 36$$

Now look at your first answer: $\frac{4}{3}x^3 -12x^2 +36x + c$. The only difference between the two is the constant term ($-36$ vs. $+c$). Since $c$ is an arbitrary constant of integration, we can simply set $c = -36$ to make the two expressions identical.

Here’s the key point: When integrating, the result always includes an arbitrary constant because the derivative of any constant is zero. Both methods are valid—they just express the family of antiderivatives in different forms, but they’re equivalent once you account for the constant.

To confirm, take the derivative of both answers:

• Derivative of $\frac{4}{3}x^3 -12x^2 +36x + c$ is $4x^2 -24x +36$, which simplifies to $(2x-6)^2$.
• Derivative of $\frac{1}{6}(2x-6)^3$ is $\frac{1}{6} \times 3(2x-6)^2 \times 2 = (2x-6)^2$, which matches perfectly.

You didn’t do anything wrong—you just have two equivalent ways to write the antiderivative!

备注：内容来源于stack exchange，提问作者88_matsy