How do I evaluate this definite integral which blows up at lower limit? I have an integral of the form
 $$\int^{\infty}_{0}{\frac{2a^2-x^{2} }{a^{2}+x^{2}}e^{\frac{-x^{2}}{b^2}}xdx}.$$
On substitution of $x^2=t$ and simplifying, I get integral of the form $$\int^{\infty}_{0}{\frac{e^{-t}}{t}dt}$$ which blows up as $ t \to 0$. Is there any way to approximate it? Is there some cutoff at the lower limit I can use?

That integral is a well known special function, the Exponential Integral . You didn't do the substitution right, though.
If $b$ is $1$, the integral is $\int_0^\infty \frac{a^2-x^2}{a^2+x^2} e^{-x^2} x dx = \int_0^\infty \frac{a^2 - t}{a^2+t} e^{-t} dt$. Then, you $u$-substitute $u=a^2+t$ and shift the bounds from $u=a^2$ to $u=\infty$ and get the integral $\int_{a^2}^\infty \frac{a^2 - (u - a^2)}{u} e^{-u} e^{a^2} du = e^{a^2} (2 a^2 \int_{a^2}^\infty \frac{e^{-u}}{u} du - \int_{a^2}^\infty e^{-u} du)$.  The second integral is simply $1- e^{-a^2}$, and the first integral is the exponential integral evaluated at $a^2$, which is finite for $a \neq 0$.
You can adjust to the case where $b \neq 1$ easily, by factoring out a $b^2$ from the numerator and denominator of the fraction and then substituting $t=\frac{x^2}{b^2}$, and then the rest is similar.
