Question

If $I={\int }_{1/e}^e\left|\log x\right|\frac{dx}{x^2}$ then $I$ equals

• A. $2$
• B. $2/e$
• C. $2(1-1/e)$
• D. $0$

Answer 1

The correct option is C $2(1-1/e)$

Substitute $\frac{1}{x}=t,$ so that
$I={\int }_e^{1/e}\left|\log \frac{1}{t}\right|(-1)dt$
$={\int }_{1/e}^e|-\log t|dt$
$={\int }_{1/e}^e\left(\log t\right)dt+{\int }_1^e\left(\log t\right)dt$
$=(t-\log t+t){]}_{1/e}^1+(t+\log t-t){]}_1^e$
$=1-\frac{1}{e}-\frac{1}{e}+e-e+1$ $=2(1-1/e).$