JEE Main 2019 (Online) 12th January Evening ShiftQuestion 1
The integral $\int_1^{e}\left\{\left(\frac xe\right)^{2x}-\left(\frac ex\right)^{x}\right\}\log_e x\,dx$ is equal to
Definite Integration
Solution
Concept. With $u=\left(\frac xe\right)^x$ we have $u'=u\ln x$, and the integrand is an exact derivative. Formulas used. (1) $u=\left(\dfrac xe\right)^x\Rightarrow\ln u=x\ln x-x\Rightarrow u'=u\ln x$Step 1 - Rewrite. $\left(\frac ex\right)^x=\frac1u$, so the integrand is $$u^2\ln x-\frac{\ln x}{u}=u\,u'-\frac{u'}{u^2}=\frac{d}{dx}\left(\frac{u^2}{2}+\frac1u\right).$$ Step 2 - Limits. At $x=e$: $u=1$. At $x=1$: $u=\frac1e$. Step 3 - Value. $$\left(\frac12+1\right)-\left(\frac{1}{2e^2}+e\right)=\frac32-e-\frac{1}{2e^2}.$$ Answer: (A)