Q.Evaluate $\displaystyle\int_{1}^{2} \frac{1}{x}\,dx$.

The antiderivative of $\tfrac1x$ is $\log|x|$; evaluate it between $1$ and $2$.

Since $\log 1 = 0$, the value is $\log 2 \approx 0.693$.

This is the reciprocal (the $n=-1$) case, whose antiderivative is $\log|x|$ (§2).

Evaluate at the limits. On $[1,2]$, $x > 0$, so $\log|x| = \log x$:

Simplify. Since $\log 1 = 0$, this is $\log 2 - 0 = \log 2 \approx 0.6931$.

Check (dual-solve): using the logarithm law $\log b - \log a = \log\tfrac{b}{a}$, the answer is $\log\tfrac{2}{1} = \log 2$ directly, the same value. (Numerically $\log 2 \approx 0.693$; here $\log$ denotes the natural logarithm, base $e$, as throughout this chapter.)

Directly by the log law: $\displaystyle\int_1^2 \tfrac1x\,dx = \log\tfrac{2}{1} = \log 2$.

Two errors recur: writing the antiderivative of $\tfrac1x$ as $\tfrac{x^0}{0}$ by misapplying the power rule (the power rule fails at $n=-1$; the antiderivative is $\log|x|$), and forgetting that $\log 1 = 0$, which would wrongly leave a stray term. Here $\log$ is the natural log ($\ln$).