Calculating limit of definite integral

Why not apply the MVT? The integral is $$F(x+1)-F(x)=\frac{F(x+1)-F(x)}{x+1-x}=F'(x_0)$$ for an $x_0$ between $x$ and $x+1$. Now $F'(x)$ certainly converges to $1$ if $x$ tends to infinity.

Note that\begin{align}\int_x^{x+1}\frac{t^2+1}{t^2+20t+8}\,\mathrm dt&=\int_x^{x+1}1\,\mathrm dt+\int_x^{x+1}\frac{-20t+7}{t^2+20t+8}\,\mathrm dt\\&=1-20\int_x^{x+1}\frac{t-\frac7{20}}{t^2+20t+8}\,\mathrm dt.\end{align}So, all that remains to be proved is that$$\lim_{x\to\infty}\int_x^{x+1}\frac{t-\frac7{20}}{t^2+20t+8}\,\mathrm dt=0.$$Can you take it from here?

For $t > 8$ you have:

$$0 \le 1 - \frac{t^2+1}{t^2+20t+8} = \frac{20t+7}{t^2+20t+8} \le \frac{20t+8}{t^2} \le \frac{21}{t}$$

Hence integrating those inequalities on $[x,x+1]$:

$$0 \le 1- \int_x^{x+1} \frac{t^2+1}{t^2+20t+8}dt \le 21 \int_x^{x+1} \frac{dt}{t} \le \frac{21}{x}$$

proving that $\lim_{x\to \infty} \int_x^{x+1} \frac{t^2+1}{t^2+20t+8}dt =1$.