JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Definite Integrals

Split the integral and handle each piece by a substitution. For ${\int }_1^e{\left(\frac{x}{e}\right)}^{2x}\ln xdx$, put $t={\left(\frac{x}{e}\right)}^{2x}$; then $\ln t=2x(\ln x-1)$ and $\frac{dt}{t}=2\ln xdx$, with $t$ running from $e^{-2}$ to $1$, so this piece is $\frac{1}{2}{\int }_{e^{-2}}^{1}dt=\frac{1}{2}(1-e^{-2})$. For ${\int }_1^e{\left(\frac{e}{x}\right)}^x\ln xdx$, put $v={\left(\frac{e}{x}\right)}^x$; then $\ln v=x(1-\ln x)$ and $dv=-v\ln xdx$, with $v$ running from $e$ to $1$, so this piece is $-{\int }_e^{1}dv=e-1$. Subtracting, $\frac{1}{2}(1-e^{-2})-(e-1)=\frac{3}{2}-e-\frac{1}{2e^2}$.