Question

If $I_1=\underset{e}{\overset{{}_{e}2}{\int }}\frac{dx}{\log x}$ and $I_2=\underset{1}{\overset{2}{\int }}\frac{e^x}{x}dx$ , then

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

Answer 1

Answer: (A)

$I=\underset{e}{\overset{e^2}{\int }}\frac{dx}{logx}$ Put log $x=z$,
$\therefore x=e^z$
$\therefore dx=e^{z}dz$
When $x=e,z=loge=1$
$x=e^2,z=log{e}^2=2loge=2$
$\therefore I_1=\underset{1}{\overset{2}{\int }}\frac{e^{z}dz}{z}=\underset{1}{\overset{2}{\int }}\frac{e^x}{z}dx=I_2$
$\therefore I_1=I_2$