Evaluate $\lim_{n\to \infty }\int_{n}^{n+1}\frac{1}{\sqrt{x^{3}+x+1}}dx$ $$\lim_{n\rightarrow \infty }\int_{n}^{n+1}\frac{1}{\sqrt{x^{3}+x+1}}dx$$
I tried to use $\int \frac{1}{\sqrt{x^{2}+a^{2}}}=\ln(x+{\sqrt{x^{2}+a^{2}}})$ but I don't get that squares from $x^{3}+x+1$.Also I took $u=x^{3}+x+1$ and $du=3x^{2}+1$  but I didn't get too far.Same for $u=\sqrt{x^{3}+x+1}$.
Some hints?
 A: Here's one: since the integrand is a positive decreasing function of $x$, one has
$$0\le \int_{n}^{n+1}\frac{1}{\sqrt{x^{3}+x+1}}\,\mathrm dx\le\int_{n}^{n+1}\frac{1}{\sqrt{n^{3}+n+1}}dx=\frac{1}{\sqrt{n^{3}+n+1}}<\frac{1}{\sqrt{n^{3}}}.$$
A: The integral $\int_1^{\infty}\frac{1}{\sqrt{x^3+x+1}}\,dx$ converges, so for every $\varepsilon$ there exists $M>0$ such that $$\int_M^{\infty}\frac{1}{\sqrt{x^3+x+1}}\,dx<\varepsilon$$
It follows that for $n>M$,
$$\int_n^{n+1}\frac{1}{\sqrt{x^3+x+1}}\,dx < \int_M^{\infty}\frac{1}{\sqrt{x^3+x+1}}\,dx<\varepsilon$$
and since the integral is positive, this proves that the limit of the l.h.s is zero.
A: If $f(x) > 0$
and
$\lim_{x \to \infty} f(x) = 0$
then
$0
\lt \int_n^{n+1} f(x) dx
\le \max(f(x))|_{x=n}^{n+1}
\to 0
$
as $n \to \infty$.
If, in addition,
$f'(x) < 0$,
then
$0
\lt \int_n^{n+1} f(x) dx
\le f(n)
\to 0
$
as $n \to \infty$.
In your case,
$f(x)
=\dfrac{1}{\sqrt{x^{3}+x+1}}
$satisfies these,
so
$ \int_n^{n+1} f(x) dx
\le f(n)
=\dfrac{1}{\sqrt{n^{3}+n+1}}
\lt \dfrac1{n^{3/2}}
$.
