How to prove that $\lim_{x \to \infty}\int_{x}^{x+1}\frac{{t^2}+1}{{t^2}+{20t}+8}dt=1$? What I did to prove that $\lim_{x \to \infty}\int_{x}^{x+1}\frac{{t^2}+1}{{t^2}+{20t}+8}dt=1$ is:
Because in the integrals ${x \to \infty}$, then the constant & ${constant*t}$ will not impact the ${t^2}$ so we can eliminate them and we get $\lim_{x \to \infty}\int_{x}^{x+1}\frac{t^2}{t^2}dt=\lim_{x \to \infty}\int_{x}^{x+1}1dt= \lim_{x \to \infty}t|_{x}^{x+1}=lim_{x \to \infty}x+1 -x = lim_{x \to \infty}1 = 1$ Does my solution correct?
 A: $$\int_x^{x+1}\frac{t^2+1}{t^2+20t+8}dt=\int_x^{x+1}\left(1-\frac{20t+7}{t^2+20t+8}\right)dt=1-\int_x^{x+1}\frac{20t+7}{t^2+20t+8}dt.$$
Now the integrand can be bounded between $0$ and $\dfrac{20}t$ and by squeezing the integral tends to $0$.
A: You intuition is exactly correct, but I'm not sure your solution has the necessary rigor. I would say that since $$\lim_{t\to\infty} \frac{t^2 + 1}{t^2 + 20t + 8} = 1$$ and since it approached this limit from below, for any $\epsilon > 0$, we can find $N > 0$ such that for $t > N$, $$1- \epsilon < \frac{t^2 + 1}{t^2 + 20t + 8} \le 1.$$ Then for $x > N$, we see $$1 - \epsilon = \int^{x+1}_x (1-\epsilon) dx\le \int_x^{x+1} \frac{t^2 + 1}{t^2 + 20t + 8} dt \le \int^{x+1}_x 1 \, dx = 1. $$ That is, for all $\epsilon > 0$, we can find $N$ such that $x >N$ implies  $$\int^{x+1}_x\frac{t^2 + 1}{t^2 + 20t + 8} dx \in[1-\epsilon, 1].$$ Thus by definition, the limit is $1$.
A: Hint:
Use the first  mean formula: That is, if $f(x)$ is continuous on $[a,b]$ then 
there exists $c\in[a,b]$ such that
$$\int_a^bf(t)dt=(b-a)f(c).$$
Therefore,
$$\int_x^{x+1}f(t)dt=f(c)=\frac{c^2+1}{c^2+20c+8},$$
where $$x\le c\le x+1.$$
When $x\to+\infty$, $c$ goes to $+\infty.$
A: In fact you can integrate a Taylor expansion because it is just inequalities in disguise and this is similar to Y.Daoust result by squeezing. The theorem that says you can integrate a Taylor expansion uses the mean value theorem as in Salahamam's answer, everything is linked.
Thus $f(t)=1-\dfrac{20}t+O\left(\frac 1{t^2}\right)$
Integrating between $[x,x+1]$ gives: $\quad 1-20\overbrace{\ln(\frac{x+1}{x})}^{O(\frac 1x)}+\overbrace{\bigg[O(\frac 1t)\bigg]_x^{x+1}}^{o(\frac 1x)}\to 1$
