QJEEM20Mada1f: JEE Main 2020 Mathematics - Class 12 JEEM20Mada1f Solution: The integral ∫ 1 2 e x . x x ( 2 + log e x ) d x equals :

Question

Verified Solution • 2025–26 Session moderate

The integral

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

equals :

Accepted Answer

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

=

\int_{1}^{2} e^{x} x^{x} [1 + (1 + \log_{e} x)] d x

=

\int_{1}^{2} e^{x} [x^{x} + x^{x} (1 + \log_{e} x)] d x

=

{[e^{x} x^{x}]}_{1}^{2}

= e²

\times

4 - e

\times

1

= 4e² - e

= e(4e - 1)

Note :

\int e^{x} (f (x) + f^{'} (x)) d x

= e^xf(x) + c

Step-by-Step Verified Solution