Just added for your curiosity.
@Song provided a nice and simple solution.
Sooner or later, you will learn that
$$\int\frac{dt}{\ln t}=\text{li}(t)$$ where appears a special function , namely the logarithmic integral function.
What is interesting is that, assuming $t>1$, the series expansion is given by
$$\text{li}(t)=\gamma +\log (t-1)+\frac{t-1}{2}-\frac{(t-1)^2}{24} +O\left((t-1)^3\right)$$ which means that, for $x$ close to $1$, using the binomial expansion
$$\int_x^{x^n}\frac{dt}{\ln t}=\log (n)+(n-1) (x-1)+\frac{(n-1)^2}{4}
(x-1)^2+O\left((x-1)^3\right)$$
For example, using $n=5$ and $x=\frac{11}{10}$, the above approximation would give $\frac{11}{25}+\log (5)\approx 2.04944$ while, using numerical integration, you would get $\approx 2.05173$.
Going a bit further, suppose that you want to compute
$$I=\int_{g(x)}^{f(x)}\frac{dt}{\ln t}$$ where you can expand the bounds as series around $x=1$ that is to say
$$f(x)=1+\sum_{i=1}^\infty a_i (x-1)^i \qquad \text{and} \qquad g(x)=1+\sum_{i=1}^\infty b_i (x-1)^i$$ you would get, as an approximation,
$$I=\log \left(\frac{a_1}{b_1}\right)+
\left(\frac{a_1}{2}+\frac{a_2}{a_1}-\frac{b_1}{2}-\frac{b_2}{b_1}\right)(x-1)+O\left((x-1)^2\right)$$