JEE Main 202228 Jul, Evening shiftMathematicsHardIntegral Calculus369× since 2002The area enclosed by the curves y=loge(x+e2),x=loge(2y)y=\log _{e}\left(x+\mathrm{e}^{2}\right), x=\log _{e}\left(\frac{2}{y}\right)y=loge(x+e2),x=loge(y2) and x=loge2x=\log _{\mathrm{e}} 2x=loge2, above the line y=1y=1y=1 is:AOption A. 2+e−loge22+\mathrm{e}-\log _{\mathrm{e}} 22+e−loge2BOption B. 1+e−loge21+e-\log _{e} 21+e−loge2COption C. e−loge2e-\log _{e} 2e−loge2DOption D. 1+loge21+\log _{e} 21+loge2CheckSkip · show solution
