Question

If $I_1=\int\limits_{e} ^{_e2}\frac{dx}{\log\,x}$ and $I_2=\int\limits_{1} ^{2}\frac{e^x}{x}dx$ , then

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

Answer 1

Answer: (A)

$I_1=I_2$

Answer 2

$I=\int\limits_{e}^{e^2} \frac{dx}{log\,x}$ Put log $x=z$, $\therefore x=e^{z}$ $\therefore dx=e^{z} dz$ When $x = e, z = log e = 1$ $x=e^{2}, z=log\,e^{2}=2 \,log e=2$ $\therefore I_{1}=\int\limits_{1}^{2} \frac{e^{z}dz}{z}=\int\limits_{1}^{2} \frac{e^{x}}{z} dx=I_{2}$ $\therefore I_{1}=I_{2}$