3 $\\int_{1}^{e} \\frac{\\ln x}{x} dx$

Explanation:

Step1: Use substitution

Let (u = \ln x), then (du=\frac{1}{x}dx). When (x = 1), (u=\ln1 = 0); when (x = e), (u=\ln e=1). The integral (\int_{1}^{e}\frac{\ln x}{x}dx) becomes (\int_{0}^{1}u du).

Step2: Integrate

According to the power - rule of integration (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), for (y = u) ((n = 1)), (\int_{0}^{1}u du=\left[\frac{u^{2}}{2}\right]_{0}^{1}).

Step3: Evaluate the definite integral

(\left[\frac{u^{2}}{2}\right]_{0}^{1}=\frac{1^{2}}{2}-\frac{0^{2}}{2}).

Answer:

(\frac{1}{2})