How do you use the integral test to determine whether $\int {\dfrac{{dx}}{{\ln x}}}
$converges or diverges from $\left[ {2,\infty } \right)$?

Hint:In order to determine the above integral converges or diverges from $\left[ {2,\infty } \right)$, consider the fact the $f\left( x \right) = \dfrac{1}{{\ln x}}$ is infinitesimal, always positive for $x > 1$ ,and decreases as the denominator increases. Since from the integral test $\int\limits_2^\infty {\dfrac{{dx}}{{\ln x}}} $can be written as $\sum\limits_{n = 2}^\infty {\dfrac{1}{{\ln n}}} $and from this we can say that $\dfrac{1}{{\ln n}} > \dfrac{1}{n}$. But as we know from the harmonic series that $\sum\limits_{n = 1}^\infty {\dfrac{1}{n}} $ is always divergent we can conclude that the
$\int\limits_2^\infty {\dfrac{{dx}}{{\ln x}}} $ will also be divergent.

Complete step by step solution:
We are given a integral $\int {\dfrac{{dx}}{{\ln x}}} $in the interval $\left[ {2,\infty } \right)$
Since, in the above integral function is $f\left( x \right) = \dfrac{1}{{\ln x}}$
Note that the above function in the interval $\left[ {2,\infty } \right)$ is
1.Infinitesimal as $\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 0$
2.$f\left( x \right) > 0$for every value of $x$greater than 1 i.e. $x > 1$ (as $\ln (1) = 0$).
3. Decreasing, as with the increase in the value of denominator the $f\left( x \right)$ will decrease.
4. $f\left( n \right) = \dfrac{1}{{\ln n}}$
So, on the basis of the integral test, the convergence of the integral $\int\limits_2^\infty
{\dfrac{{dx}}{{\ln x}}} $is equal to the convergence of the series $\sum\limits_{n = 2}^\infty
{\dfrac{1}{{\ln n}}} $
Now, if we look on the above carefully, we can easily demonstrate that the $\ln n < n$
So that $\dfrac{1}{{\ln n}} > \dfrac{1}{n}$
And as know that the harmonic which says :
$\sum\limits_{n = 1}^\infty {\dfrac{1}{n}} $is always divergent.
Now we can also conclude that the $\sum\limits_{n = 2}^\infty {\dfrac{1}{{\ln n}}} $ will also be divergent by directly comparing with above.
Hence, also $\int\limits_2^\infty {\dfrac{{dx}}{{\ln x}}} $ is divergent .
Formula:
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} $
\[{\cos ^2}x + {\sin ^2}x = 1\]
$\int {f(x)g'(x)dx = f(x)g(x) - \int {f'(x)g(x)dx} } $
Additional Information:
Different types of methods of Integration:
Integration by Substitution
Integration by parts

Note:
1.Use standard formula carefully while evaluating the integrals.
2. Indefinite integral=Let $f(x)$ be a function .Then the family of all its primitives (or antiderivatives)
is called the indefinite integral of $f(x)$ and is denoted by $\int {f(x)} dx$
3.The symbol $\int {f(x)dx} $ is read as the indefinite integral of $f(x)$with respect to x.