JEE Main 2022 — Mathematics Calculus

Given,

$$f(x)=\frac{|{x}^{3}+x|}{({e}^{x|x|}+1)}dx$$f(x)=(ex∣x∣+1)∣x3+x∣​dx

and $\displaystyle {\int }_{-2}^{2}f(x)dx={\int }_{0}^{2}(f(x)+f(-x))dx$∫−22​f(x)dx=∫02​(f(x)+f(−x))dx

$$={\int }_{0}^{2}(\frac{|{x}^{3}+x|}{({e}^{x|x|}+1)}+\frac{|-{x}^{3}-x|}{({e}^{-x|-x|}+1)})dx$$=∫02​((ex∣x∣+1)∣x3+x∣​+(e−x∣−x∣+1)∣−x3−x∣​)dx

$$={\int }_{0}^{2}(\frac{|{x}^{3}+x|}{({e}^{x|x|}+1)}+\frac{|{x}^{3}+x|}{({e}^{-x|x|}+1)})dx$$=∫02​((ex∣x∣+1)∣x3+x∣​+(e−x∣x∣+1)∣x3+x∣​)dx

$$={\int }_{0}^{2}(\frac{{x}^{3}+x}{({e}^{{x}^{2}}+1)}+\frac{{x}^{3}+x}{({e}^{-{x}^{2}}+1)})dx$$=∫02​((ex2+1)x3+x​+(e−x2+1)x3+x​)dx

$$I={\int }_{0}^{2}(\frac{{x}^{3}+x}{1+{e}^{{x}^{2}}}+\frac{{e}^{{x}^{2}}({x}^{3}+x)}{1+{e}^{{x}^{2}}})dx$$I=∫02​(1+ex2x3+x​+1+ex2ex2(x3+x)​)dx

$$={\int }_{0}^{2}({x}^{3}+x)dx$$=∫02​(x3+x)dx

$$={[\frac{{x}^{4}}{4}+\frac{{x}^{2}}{2}]}_{0}^{2}$$=[4x4​+2x2​]02​

$$=4+2=6$$=4+2=6