Question

The value of the integral $\underset{-2}{\overset{2}{\int }}\frac{|x^3+x|}{\left(e^{x|x|}+1\right)}dx$ is equal to:

• A. $5e^2$
• B. $3e^{-2}$
• C. $4$
• D. $6$

Answer 1

Answer: (D)

$f(x)=\frac{|x^3+x|}{\left(e^{x|x|}+1\right)}dx$
$\underset{-2}{\overset{2}{\int }}f(x)dx={\int }_0^2(f(x)+f(-x))dx$
$=\underset{0}{\overset{2}{\int }}\left(\frac{|x^3+x|}{\left(e^{x|x|}+1\right)}+\frac{|-x^3-x|}{\left(e^{-x|-x|}+1\right)}\right)dx$
$=\underset{0}{\overset{2}{\int }}\left(\frac{|x^3+x|}{\left(e^{x|x|}+1\right)}+\frac{|x^3+x|}{\left(e^{-x|x|}+1\right)}\right)dx$
$=\underset{0}{\overset{2}{\int }}\left(\frac{x^3+x}{\left(e^{x^2}+1\right)}+\frac{x^3+x}{\left(e^{-x^2}+1\right)}\right)dx$
$I=\underset{0}{\overset{2}{\int }}\left(\frac{x^3+x}{1+e^{x^2}}+\frac{e^{x^2}\left(x^3+x\right)}{1+e^{x^2}}\right)dx$
$=\underset{0}{\overset{2}{\int }}\left(x^3+x\right)dx$
$={\left[\frac{x^4}{4}+\frac{x^2}{2}\right]}_0^2$
$=4+2=6$