Question

$\underset{1}{\overset{2}{\int }}\log xdx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_1^2\log x\mathit{d}x.\mathrm{Then}, \\ I={\int }_1^{2}1\log x\mathit{d}x \\ \mathrm{Integrating}\mathrm{by}\mathrm{parts} \\ \Rightarrow I={\left[x\log x\right]}_1^2-{\int }_1^2\frac{1}{x}x\mathit{}\mathit{d}x \\ \Rightarrow I={\left[x\log x\right]}_1^2-{\int }_1^2\mathit{d}x \\ \Rightarrow I={\left[x\log x\right]}_1^2-{\left[x\right]}_1^2 \\ \Rightarrow I=2\log 2-2+1 \\ \Rightarrow I=2\log 2-1 \end{gathered}$$