Compute $\int_0^1 \frac {x - 1} {\ln x} \, dx$ $$\int_0^1 \frac {x - 1} {\ln x} \, dx =\, ? $$
I have not any idea how start it.  Should I use change of variable? Or partial integration? $\frac 1 {\ln x}=u \rightarrow (x-1)dx=dv$  is it true? I cannot find conclusion.
 A: Actually, the natural change of variable yields the result... Using $x=\mathrm e^{-t}$, one sees that the integral to be computed is $F(2)$, where, for every $a\gt1$,
$$
F(a)=\int_0^\infty\frac{\mathrm e^{-t}-\mathrm e^{-at}}t\,\mathrm dt.
$$
The function $F$ is smooth on $(1,+\infty)$ and, for every $a\gt1$,
$$
F'(a)=\int_0^\infty\frac{\partial}{\partial a}\left(\frac{\mathrm e^{-t}-\mathrm e^{-at}}t\right)\,\mathrm dt=\int_0^\infty\mathrm e^{-at}\,\mathrm dt=\frac1a.
$$
Since $\lim\limits_{a\to1}F(a)=0$, $F(a)=\log a$ for every $a\gt1$. In particular,
$$\int_0^1 \frac {x - 1} {\ln x} \,\mathrm dx =\log2.
$$
A: Okay so I figured it out by digging in my old books.
So there is one infamous represention of the natural logarithm as
$$\ln(x) = -\lim_{n->0} \int_n^{\infty}\frac{e^{-xt}-e^{-t}}{t}dt$$
In your case if we let $u=-\ln(x)$ the integral becomes
$$-\int_0^{\infty}\frac{e^{-2u}-e^{-u}}{u}du = \ln(2)$$
Equality holds as improper integral evaluation is defined as the limit of the proper.
To proove the first statement there are 2 major points:
1) $\ln(1) = 0$ and for $x=1$ the limit is 
$$-\lim_{n->0} \int_n^{\infty}\frac{e^{-xt}-e^{-t}}{t}dt=-\lim_{n->0} \int_n^{\infty}\frac{e^{-t}-e^{-t}}{t}dt = \lim_{n->0} \int_n^{\infty}0dt = 0$$
2) We proove that their derivatives with respect to $x$ are euqal:
$$\frac{d\ln(x)}{dx} = \frac{d\left[-\lim_{n->0} \int_n^{\infty}\frac{e^{-xt}-e^{-t}}{t}dt\right]}{dx}$$
3)Since both of these are true $\forall x > 0$ by the fundamental theorem of calculus it follows that their equal up to a constant. Since we also proved and 1) it follows that they are exactly euqal. Write me down if you need a proof of point 2).
