How can I evaluate $\int_{-\infty}^{\infty}\frac{e^{-x^2}(2x^2-1)}{1+x^2}dx$? How can I solve this integral: $$\int_{-\infty}^{\infty}\frac{e^{-x^2}(2x^2-1)}{1+x^2}dx.$$ Can I solve this problem using the Laplace transform? How can I do this?

Writing $2x^2-1$ as $2x^2 + 2 - 3$, the integral simplifies to
$$\begin{align*}
\int_{-\infty}^{\infty}\exp(-x^2)\frac{2x^2-1}{1+x^2}\mathrm dx
&= 2\int_{-\infty}^{\infty}\exp(-x^2)\mathrm dx
-3 \int_{-\infty}^{\infty}\exp(-x^2)\frac{1}{1+x^2}\mathrm dx\\
&= 2\sqrt{\pi} 
- \frac{3}{2\pi}\int_{-\infty}^{\infty}\sqrt{\pi}\exp(-\omega^2/4)
\pi\exp(-|\omega|)\mathrm d\omega\\
&= 2\sqrt{\pi} 
- 3\sqrt{\pi}\int_{0}^{\infty}\exp(-\omega^2/4)
\exp(-\omega)\mathrm d\omega\\
&= 2\sqrt{\pi} 
- 6\pi e \int_{0}^{\infty}\frac{1}{\sqrt{2}\sqrt{2\pi}}
\exp\left(-\frac{(\omega + 2)^2}{2\cdot(\sqrt{2})^2}\right)
\mathrm d\omega\\
\end{align*}$$
where in the second step, we have used a well-known result
(see e.g. the answer by bgins and the links therein) on the first
integral, and applied the inner-product form of Parseval's theorem 
to convert the second integral to the integral
of the product of the Fourier transforms $\sqrt{\pi}\exp(-\omega^2/4)$
and $\pi\exp(-|\omega|)$ of $\exp(-x^2)$ and
$(1+x^2)^{-1}$ respectively,   while the last step follows upon 
completing the square in the exponent and writing the 
integrand as the  probability density function of a 
normal random variable with mean $-2$ and variance  $2$.
Hence we have that
$$\begin{align*}
\int_{-\infty}^{\infty}\exp(-x^2)\frac{2x^2-1}{1+x^2}\mathrm dx
&= 2\sqrt{\pi} -6\pi e \Phi(-2/\sqrt{2})\\
&= 2\sqrt{\pi} -6\pi e Q(\sqrt{2})
\end{align*}$$
where $\Phi(\cdot)$ is the cumulative  standard normal
distribution function and $Q(x) = 1-\Phi(x)$ is its complement.
Since $\text{erfc}(x) =  2Q(x\sqrt{2})$, the value can also be 
expressed
as $2\sqrt{\pi} - 3e\pi \text{erfc}(1)$ as in the answers by Bruno and bgins.
