QJEEM22Maa8c7: JEE Main 2022 Mathematics - Class 12 JEEM22Maa8c7 Solution: The value of the integral is equal to :

$I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{x|x|}} + 1}}dx}$ ..... (i)

$I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{ - x|x|}} + 1}}dx}$ ..... (ii)

$2I = \int\limits_{ - 2}^2 {|{x^3} + x|dx}$

$2I =2 \int\limits_0^2 {({x^3} + x)dx}$

$I = \int\limits_0^2 {({x^3} + x)dx}$

$= \left. {{{{x^4}} \over 4} + {{{x^2}} \over 2}} \right]_0^2$

$= \left( {{{16} \over 4} + {4 \over 2}} \right) - 0$

$= 4 + 2 = 6$