I am reading something on the Laplace method of integrals and I don't understand part of it's argument. It gives the integral
$$\int_{-3}^4 e^{-\lambda x^2}\log(1+x^2)dx$$
and finding the leading term of asymptotics for $\lambda\to\infty$. It first argues
"Since only small $|x|$, such that $|x| \sim \frac{1}{\sqrt{\lambda}} \ll 1$ are important,
$$\log(1+x^2)\sim x^2$$
$$I(\lambda)\sim \int_{-3}^4 e^{-\lambda x^2}x^2 \, dx \sim \int_{-\infty}^{\infty} e^{-\lambda x^2}x^2 \, dx \sim 2\int_{0}^{\infty} e^{-\lambda x^2}x^2 \, dx." $$
I'm really not following that argument at all. Any help would be appreciated.
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