关于用换元法求解∫1/x³积分的疑问及正确推导步骤请求

Hey there! Let's start by clearing up a quick mix-up first: the result $\frac{x^4}{4}$ you thought was correct? That's actually the integral of $x^3$, not $\frac{1}{x^3}$. Your u-substitution result $-\frac{1}{2}x^{-2}$ is totally right—it matches exactly what you'd get using the power rule directly!

I'll walk through the steps (following your approach) to confirm this, and show where the confusion might have come in:

First, let's restate the integral we're solving:
$$
\int{\frac{1}{x^3}dx}
$$

Your u-substitution steps (all correct, by the way!) #

1. Set $u = x^3$, which means $x = u^{1/3}$
2. Differentiate to find $dx$: $du = 3x^2 dx$, so rearranged gives $dx = \frac{1}{3x^2} du$
3. Substitute these into the original integral:
   $$
   \int{\frac{1}{u} \times \frac{1}{3x^2} du}
   $$
4. Replace $x^2$ with $u^{2/3}$ (since $x = u^{1/3}$, squaring both sides gives $x^2 = u^{2/3}$):
   $$
   \int{\frac{1}{u} \times \frac{1}{3u^{2/3}} du}
   $$
5. Combine the terms using exponent rules ($u^{-1} \times u^{-2/3} = u^{-5/3}$), and factor out the constant:
   $$
   \frac{1}{3} \int{u^{-5/3} du}
   $$
6. Apply the power rule for integrals ($\int{u^n du} = \frac{u^{n+1}}{n+1} + C$ for $n \neq -1$):
   $$
   \frac{1}{3} \times \frac{u^{-2/3}}{-2/3} + C
   $$
7. Simplify the constants:
   $$
   \frac{1}{3} \times \left(-\frac{3}{2}\right) u^{-2/3} + C = -\frac{1}{2} u^{-2/3} + C
   $$
8. Substitute back $u = x^3$:
   $$
   -\frac{1}{2} (x³⁾{-2/3} + C = -\frac{1}{2} x^{-2} + C = -\frac{1}{2x^2} + C
   $$

Quick power rule check to confirm #

Using the power rule directly on $\int{x^{-3} dx}$:
$$
\frac{x^{-3+1}}{-3+1} + C = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C
$$

Same result! So your u-substitution method worked perfectly—you didn't make a mistake at all. The only mix-up was confusing the integral of $x^3$ (which is $\frac{x^4}{4} + C$) with the integral of $\frac{1}{x^3}$. Nice work working through the substitution step-by-step!

备注：内容来源于stack exchange，提问作者Robert Barnett