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.
 A: 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, 
$$\begin{align}\int_x^{x^2}\frac{1}{\log(t)}\,dt&=\int_x^{x^2}\left(\frac{1}{\log(t)}-\frac{1}{t-1}\right)\,dt+\int_x^{x^2}\frac{1}{t-1}\,dt\tag1
\end{align}$$
A: 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
$$\lim_{x\to 1}f(x)=\ln 2.$$
A: Use the fact that $\ln$ is a concave function, and thus, for $1\leq x\leq t\leq x^2$,
$$2(t-1)\frac{\ln(x)}{(x-1)(x+1)}\leq\ln(t)\leq t-1$$
Invert this:
$$\int_x^{x^2}\frac1{t-1}~\mathrm dt\leq\int_x^{x^2}\frac1{\ln(t)}~\mathrm dt\leq\frac{(x-1)(x+1)}{2\ln(x)}\int_x^{x^2}\frac1{t-1}~\mathrm dt$$
These are easy to integrate:
$$\int_x^{x^2}\frac1{t-1}~\mathrm dt=\ln(x+1)$$
And so by the squeeze theorem, one can show that
$$\lim_{x\to1}\int_x^{x^2}\frac1{\ln(t)}~\mathrm dt=\ln(2)$$
Noting the right integral mess can be fixed up with the first inequality applied one more time and using the squeeze theorem.
A: $\frac{1}{\ln(t)}=\frac{t}{t\ln(t)}=\frac{(t-1)+1}{t\ln(t)}=\frac{t-1}{t\ln(t)}+\frac{1}{t\ln(t)}$
Let $f : t \mapsto \frac{t-1}{t\ln(t)}$ and $g : t \mapsto \frac{1}{t\ln(t)}$.


*

*$f$ can be continuously extended in 1, thus $\int f(t) ~\mathrm{d}t$ converges. Then, $\int_x^{x^2} f(t) ~\mathrm{d}t \underset{x=1}{\rightarrow}0$.

*$g(t) = \frac{d}{dt}(\ln(\ln(t))$, and $\ln(\ln(x^2))-\ln(\ln(x))=\ln(2\ln(x))−\ln(\ln(x))=\ln(2)$


Conclusion: the limit is $\ln(2)$.
A: This seems like a good problem for generalization, since the proof for the general result might actually be easier. Suppose $f$ is continuous near $1,$ $f(1)=0,$ and $f'(1)\ne 0.$ Then
$$\tag 1 \lim_{x\to 1} \int_x^{x^2} \frac{1}{f(t)}\,dt = \frac{\ln 2}{f'(1)}.$$
With $f(t)= \ln t,$ we get the answer of $\ln 2$ for the given problem.
Rough sketch of proof of $(1)$:
$$\frac{1}{f(t)} = \frac{1}{t-1}\cdot \frac{t-1}{f(t)}.$$
Now the second fraction has limit $1/f'(1)$ as $t\to 1.$ So given $\epsilon>0,$ this fraction lies between the constants $(1-\epsilon)/f'(1)$ and $(1+\epsilon)/f'(1)$ for $t$ close enough to  $1.$ Move those constants outside the integral, evaluate the easy integral remaining, and the result follows.
A: Setting $t = e^s$ we get
$$
f(x) 
= \int_{\ln x}^{2 \ln x} \frac{e^s}{s}\ ds
= \int_{\ln x}^{2 \ln x} \frac{1+s+O(s^2)}{s}\ ds
= \int_{\ln x}^{2 \ln x} \left(\frac{1}{s}+1+O(s)\right) \ ds \\
= (\ln(2 \ln x) - \ln(\ln x)) + (2\ln x - \ln x) + O((\ln x)^2) \\
= \ln 2 + \ln x + O((\ln x)^2) 
\to \ln 2
$$
as $x \to 1$.
A: 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 $\ln2$.
