Question

The value of the integral ${\int }_{-a}^a\frac{{xe}^{x^2}}{1+x^2}dx$ is

• A. $e^{a^2}$
• B. $0$
• C. $e^{{-a}^2}$
• D. $a$

Answer 1

The correct option is B $0$

Since, $\frac{{xe}^{x^2}}{1+x^2}$ is an odd function.
$\therefore {\int }_{-a}^a\frac{{xe}^{x^2}}{1+x^2}dx=0$