<s>[INST] Calculating limit of definite integral I need to calculate:
$$
\lim_{x\to \infty} \int_x^{x+1} \frac{t^2+1}{t^2+20t+8}\, dt
$$ The result should be $1$.
Is there a quicker way than calculating the primitive function?
I thought about seperating to $\int_0^{x+1} -\int_0^x$ but still can't think of the solution. 
 [/INST] Use the estimate:
$$\frac{t-10}{t+10}<\frac{t^2+1}{t^2+20t+8}<\frac t{t+10}, \ t>1 \Rightarrow \\
\int_x^{x+1} \frac{t-10}{t+10}dt<I(x)<\int_x^{x+1} \frac t{t+10}dt \Rightarrow \\
1-20\ln \frac{x+11}{x+10}<I(x)<1-10\ln \frac{x+11}{x+10} \Rightarrow \\
\lim_{x\to\infty} \left(1-20\ln \frac{x+11}{x+10}\right)\le \lim_{x\to\infty} I(x) \le \lim_{x\to\infty} \left(1-10\ln \frac{x+11}{x+10}\right) \Rightarrow \\
\lim_{x\to\infty} I(x)=\lim_{x\to\infty} \int_x^{x+1}\frac{t^2+1}{t^2+20t+8}dt=1.$$
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