# Integral with delta-function in denominator

I have the integral $$I=\int\limits_{-\infty}^{+\infty} dx\,\frac{f(x)}{a(x)+[b(x)+\delta(x)]^2},$$ where $f(x)$, $a(x)$, and $b(x)$ are some smooth functions, and $\delta(x)$ is the Dirac delta-function. Moreover, $f(0)=0$.

Is it correct that we can just drop $\delta(x)$ from the integrand, so $$I=\int\limits_{-\infty}^{+\infty} dx\,\frac{f(x)}{a(x)+b^2(x)},$$ and this can be done even if $f(0)\neq0$?

The integral as it stands isn't mathematically rigorous, but we could replace $\delta$ with a sequence of function $u_k$ converging to $\delta$ in the space of distributions and take limits.
If we for example take $u_k(x) = \frac{1}{2\epsilon} \chi_{[-\epsilon, +\epsilon]}(x)$ then we have $$I_k = \int_{|x|>\epsilon} \frac{f(x)}{a(x)+b(x)^2} \, dx + \int_{|x|<\epsilon} \frac{f(x)}{a(x)+(b(x)+\frac{1}{2\epsilon})^2} \, dx \\ = \int_{|x|>\epsilon} \frac{f(x)}{a(x)+b(x)^2} \, dx + \int_{|x|<\epsilon} \frac{4\epsilon^2 f(x)}{4 \epsilon^2 a(x)+(2\epsilon b(x)+1)^2} \, dx \\ \to \int_{\mathbb R} \frac{f(x)}{a(x)+b(x)^2} \, dx$$ unless $f(x) \to \infty$ too fast as $x \to 0$.