Question

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

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

Answer 1

Answer: (B)

Since, $\frac{x{e}^{x^2}}{1+x^2}$ is an odd function.

$\therefore \underset{-a}{\overset{a}{\int }}\frac{x{e}^{x^2}}{1+x^2}dx=0$