Q.Evaluate $\displaystyle\int_{1}^{e} \frac{\log x}{x}\,dx$ using substitution.

Here $\tfrac1x$ is the derivative of $\log x$, so substitute $u=\log x$ and change the limits.

Recognise the $\log x$ pattern of §4 (a $\tfrac1x$ factor multiplying a function of $\log x$).

Choose the substitution. Let $u = \log x$. Then $du = \dfrac{1}{x}\,dx$, which is precisely the $\dfrac{1}{x}\,dx$ present in the integrand.

Change the limits. When $x = 1$: $u = \log 1 = 0$. When $x = e$: $u = \log e = 1$ (natural log). So the limits $1 \to e$ become $0 \to 1$.

Rewrite and evaluate in $u$.

$\int_{1}^{e} \frac{\log x}{x}\,dx = \int_{0}^{1} u\,du = \left[\frac{u^2}{2}\right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}.$ …

A common error is forgetting $\log e = 1$ and $\log 1 = 0$ when changing the limits, which corrupts the new integral. Another is keeping the $x$-limits $1$ and $e$ afte …