conditions for improper integral convergent 
Quiz : let $p$ be constant, if the improper integral $ \int_{0}^{1}  \frac{\ln x}{x^p(1-x)^{1-p}} \ dx$  convergent, then what's the range of p ?
Actually I'm clueless,

*

*but i know when $x$ tends to $0$  and $p>0$ ,so that $lnx$ goes to infinity and $\frac{1}{x^p}$ tends to infinity too, so $x$ =1 is a improper point


*when $x$ goes to 1 and $p<1$ ,so that $\frac{1}{(1-x)^{1-p}}$ tends to infinity,so $x=1$ is a improper point too


*no further then i don't know how to work out this question ,so confused
 A: Too long for a comment
The integrand has "dangerous" points only on the boundaries of the interval (at $x=0$ and $x=1$), so we should investigate the integral's behavior only near these points. Setting $a$ such that $0<a<1$, it is enough to study convergence of $\int_0^a$ and $\int_a^1$.
For $x=0$, integrating by part, we can make the following estime:
$$\Big|\int_0^a\frac{\ln x}{x^p(1-x)^{1-p}} dx\Big|<\int_0^a\frac{\ln x}{x^p}dx\,\cdot\Big|\max\frac{1}{(1-x)^{1-p}}\Big|_{x\in[0;a]}$$
$$=\Big|\max\frac{1}{(1-x)^{1-p}}\Big|_{x\in[0;a]}\cdot\Big|\frac{1}{1-p}\Big|\cdot\Big|\Big(\,x^{1-p}\ln x\,\Big|_0^a-\int_0^ax^{-p}dx\Big)\Big|$$
The last expression is finite at $p<1$.
In the same way, for $x=1$
$$\Big|\int_a^1\frac{\ln x}{x^p(1-x)^{1-p}} dx\Big|<\Big|\int_a^1\frac{\ln x}{(1-x)^{1-p}}dx\Big|\,\cdot\Big|\max\frac{1}{x^p}\Big|_{x\in[a;1]}$$
Making a change of variable $1-x=t$ and using the series $\ln(1-t)=-t-\frac{t^2}{2}+O(t^3)$
$$\Big|\int_a^1\frac{\ln x}{x^p(1-x)^{1-p}} dx\Big|<\Big|\int_0^{1-a}\frac{\ln (1-t)}{t^{1-p}}dt\Big|\,\cdot\Big|\max\frac{1}{x^p}\Big|_{x\in[a;1]}$$
$$=\int_0^{1-a}\frac{t+\frac{t^2}{2}+..}{t^{1-p}}dt\cdot\Big|\max\frac{1}{x^p}\Big|_{x\in[a;1]}=\Big(\frac{t^{p+1}}{p+1}\Big|_0^{1-a}+\frac{t^{p+2}}{p+2}\Big|_0^{1-a}+..\Big)\cdot\Big|\max\frac{1}{x^p}\Big|_{x\in[a;1]}$$
This expression is finite for $p>-1$.
Therefore, as @Daniel Clarke mentioned in the comment, the integral converges at $\,\,-1<p<1$
A: There are two singular points $0$ and $1$. Therefore we study two integrals
over $(0,1/2]$ and over  $[1/2,1).$ For $x\to 1^-$ we have $\ln x=\ln[1-(1-x)]\approx -(1-x).$ Hence $${\ln x\over x^p(1-x)^{1-p}} \approx -(1-x)^{-p},\quad x\to 1^-$$ Thus $p>-1$ is the necessary and sufficient condition for integrability of the integral over $[1/2,1).$ At $x\to 0^+$ the numerator tends to $-\infty$ and the denominator behaves like $x^p.$ Therefore the condition $p<1$ is necessary for integrability of the integral over $(0,1/2).$
Actually the condition is also sufficient as we can calculate the antiderivative explicitly (we ignore $(1-x)^{p-1}\approx 1$ when $x\to 0^+$)
$$\int {\ln x\over x^p}\,dx ={1\over 1-p}x^{1-p}\,\ln x-{1\over (1-p)^2}\,x^{1-p}$$ and the limit of the antiderivative at $x\to 0^+$ is equal $0$ for $p<1.$
