How to solve this improper integral using Simpson's rule

Question

I'm trying to calculate value of $\int_{0}^{\infty}{e^{-2x}(1+e^{-x})}dx$ using Simpson 1/3 rule. I know the solution is substitution $x$ by $g(t)$ but every $g(t)$ I've tested the result was incorrect.

Answer 1

Substitute $$x\to \frac{1-t}{t+1};\;dx=-\frac{2dt}{(t+1)^2};\;t\to \frac{1-x}{x+1}$$ Integral becomes $$\int_1^{-1} \frac{(-2) e^{-\frac{2 (1-t)}{t+1}} \left(e^{-\frac{1-t}{t+1}}+1\right)}{(t+1)^2} \, dt=\int_{-1}^1 \frac{2 e^{\frac{2 (t-1)}{t+1}} \left(e^{\frac{t-1}{t+1}}+1\right)}{(t+1)^2} \, dt$$ Notice that $\underset{t\to -1}{\text{lim}}f(t)=0$ so you can set $f(-1)=0$