test for convergence of improper integral1 $$\int_0^1 {x^n\log x\over(1+x)^2} \, dx$$
I tried something using practical test, but not much progress. 
I see that the integral becomes improper for $x=0$, May be we need to apply the Practical comparison test ?
I need some hint please.
 A: It's simple to see that the integral
$$\int_0^1|\log x|\, dx$$
is convergent so conclude with the inequalities
$$\frac{x^n|\log x|}{(1+x)^2}\leq |\log x|,\quad 0<x\leq1, \, n\geq0$$
and
$$-\frac{x^n\log x}{(1+x)^2}\sim_0-x^n\log x\geq\frac{1}{x^{-n}}\quad 0<x\leq\frac{1}{e}, \, n<0$$
A: Hint: $$\lim_{x\to 0+} x^n\ln x=0,\quad n> 0$$ and $$\left|\int_0^1\ln x\,dx\right|<\infty.$$
A: A related technique. Note that, 
$$ x\sim 0  \implies \frac{x^n\ln(x)}{(1+x^2)}\sim x^n\ln(x).$$
Now, the integral
$$ \int_{0}^{1}x^n\ln(x) dx = \frac{x^{n+1}\ln(x)}{n+1}\Big|_{x=0}^{x=1}-\frac{1}{n+1}\int_{0}^{1} x^{n+1}dx=-\frac{1}{n+1}\int_{0}^{1} x^{n+1}dx. $$
Now, do you know for what $n$ the last integral converge?
A: This is in no sense an improper integral.
$$
x\log x = \frac{\log x}{1/x},
$$
and by L'Hopital's rule this goes to $0$ as $x\downarrow0$.  Therefore $x^n\log x\to0$ as $x\downarrow0$.  The denominator stays far from $0$ over the whole interval.
So the function is bounded, and it's integrated over a bounded interval, so there is no "impropriety".  No test for convergence is needed.
