Find $\lim_{x\to 1}\int_{x}^{x^2}\frac{1}{\ln {t}}\mathrm dt$.

Find $\lim_{x\to 1}f(x)$, where $$f(x) = \int_{x}^{x^2}\frac{1}{\ln {t}}\mathrm dt$$

I tried splitting it into two integrals, one from 1 to $x^2$ and the other one from $x$ to $1$. Doesn't matter how I split it, I got zero. Wolfram Alpha has some kind of a different, more sophisticated answer with functions we didn't cover at the lecture.

For $t>0$: $$\frac{t-1}t\leq\ln t\leq t-1$$ so $$\int_x^{x^2}\frac{1}{t-1}dt\leq\int_x^{x^2} \frac{1}{\ln t}\,dt\leq \int_x^{x^2}\dfrac{t}{t-1}dt$$ with limit $x\to1$ you have answer $\ln2$.

Substituting $t=e^u$, $\mathrm dt=e^u\mathrm du$, we have $$ f(x)=\int_{\ln x}^{2\ln x}\frac{e^u\mathrm du}{u}.$$ For $x>1$ we then have $1\le e^u\le x^2$, hence $$ \int_{\ln x}^{2\ln x}\frac{\mathrm du}{u}\le f(x)\le x^2\int_{\ln x}^{2\ln x}\frac{\mathrm du}{u}$$ (and similar for $x<1$) with a now well-known integral: $\int \frac{\mathrm du}u=\ln |u|+C$. We conclude that $$ \ln 2\le f(x)\le x^2\ln 2$$ and so $$\lim_{x\to 1}f(x)=\ln 2.$$

HINT:

Write $\frac{1}{\log(t)}=\left(\frac{1}{\log(t)}-\frac{1}{t-1}\right)+\frac{1}{t-1}$. Then note that $\frac{1}{\log(t)}-\frac{1}{t-1}$ has a removable discontinuity at $t=1$, which once removed, is analytic in a neighborhood of $t=1$.

Then write,

$$\begin{align}\int_x^{x^2}\frac{1}{\log(t)}\,dt&=\int_x^{x^2}\left(\frac{1}{\log(t)}-\frac{1}{t-1}\right)\,dt+\int_x^{x^2}\frac{1}{t-1}\,dt\tag1 \end{align}$$