The integral ∫12e^x. x^x 2+log_e x dx equals :

Correct answer: (C) e4e-1. Let y=ex^x lny=x1+lnx 1/ydy/dx=2+lnx ⇒ dy=ex^x2+lnx dx ∫12e^x.x^x2+log_exdx =y12 =ex^x12 =4e^2−e

Q26 / 75 · JEE Main 2020 (06 Sep Shift 2) · Mathematics

The integral $\int_{1}^{2} e^{x} . x^{x} (2 + \log_{e} x) dx$ equals :

Let $y = {(ex)}^{x}$

$l ny = x [1 + l nx]$

$\frac{1}{y} \frac{dy}{dx} = (2 + l nx)$

$\Rightarrow$ $dy = {(ex)}^{x} (2 + l nx) dx$

$\int_{1}^{2} e^{x} . x^{x} (2 + \log_{e} x) dx$

$= {(y)}_{1}^{2}$

$= {({(ex)}^{x})}_{1}^{2}$

$= 4 e^{2} - e$

Hence, the correct answer is option: 'C'.