Prove that $\lim_{x\to 1^{+} } \int_{x}^{x^{3}}\frac{1}{\ln t}\, dt=\ln3$.

Substitute $t=x^v$, then we obtain $$\begin{align*} I(x)&=\int_{x}^{x^3}\frac1{\ln t}\ dt\\&=\int_1^3 \frac{x^{v}}{v} \ dv. \end{align*}$$ Now, we can see $\lim_{x\to 1^+}I(x)=\ln 3$ by Lebesgue's dominated convergence theorem; for every sequence $x_n>1$ converging to $1$, we have that $$\begin{align*} I(x_n)&=\int_1^3 \frac{x_n^{v}}{v} \ dv\\&\xrightarrow{n\to \infty}\int_1^3 \frac 1 v\ dv=\ln 3 \end{align*}$$ by LDCT. (For $1<x\le 2$, $0\le \frac{x^v}{v}\le \frac{2^v}{v}$.)

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, $$I=\log \left(\frac{a_1}{b_1}\right)+ \left(\frac{a_1}{2}+\frac{a_2}{a_1}-\frac{b_1}{2}-\frac{b_2}{b_1}\right)(x-1)+O\left((x-1)^2\right)$$