Practicing Success
If $I_1=\int\limits_{e}^{e^2}\frac{dx}{\log x}$ and $I_2=\int\limits_{1}^{2}\frac{e^x}{x}dx$, then |
$I_1=I_2$ $2I_1=I_2$ $I_1=2I_2$ none of these |
$I_1=I_2$ |
$I_2=\int\limits_{1}^{2}\frac{e^x}{x}dx$ let $y=e^x⇒dy=e^xdx$ so $x=\log y$ as $x → 1, y → e$ $x → 2, y → e^2$ $I_2=\int\limits_{e}^{e^2}\frac{dy}{\log y}$ $⇒I_2=\int\limits_{e}^{e^2}\frac{dx}{\log x}=I_1$ |