Why is the integral of 1/x equal to the natural logarithm of x?

The fundamental reason is the derivative of $\ln x$ is $1/x$, but this can be seen as a definitional choice. **Approach 1 (Natural log defined as integral):** Define $\ln x = \int_1^x \frac{1}{t} dt$. Then by the fundamental theorem, $\frac{d}{dx} \ln x = \frac{1}{x}$. This is actually how $\ln x$ is often defined in advanced calculus. **Approach 2 (Why $x^{-1}$ is special):** For any power $x^n$, the derivative is $nx^{n-1}$. The integral of $x^n$ is $\frac{x^{n+1}}{n+1} + C$, except when $n = -1$ because we'd divide by zero. The gap in the power rule is filled by the logarithm. **Approach 3 (From exponential):** If $y = \ln x$, then $e^y = x$. Differentiating: $e^y \frac{dy}{dx} = 1$, so $\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$. For other powers like $x^{-2}$, the power rule works fine: $\int x^{-2} dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C$. Only $n = -1$ is exceptional because it's the only power that integrates to a logarithm.