Question

If $I_1={\int }_e^{e^2}\frac{dx}{logx}$ and $I_2={\int }_1^2\frac{e^x}{x}dx,$ then

• A. $I_1=I_2$
• B. $2I_1=I_2$
• C. $I_1=2I_2$
• D. $none$

Answer 1

The correct option is A $I_1=I_2$

Put log $x=y\Rightarrow x=e^y\Rightarrow dx=e^y.dy$
The limit of y are 1, 2
${\int }_e^{e^2}\frac{1}{logx}dx={\int }_1^2\frac{1}{y}.e^{y}dy={\int }_1^2\frac{e^x}{x}dx\therefore I_1=I_2$