$$\int_0^1{{x^n\log x}\over {(1+x)^2}} $$ My Attempt :
The integrand is unbounded for $x=0$. so,
Using $\mu-Test$ , the integral will be convergent if $0<\mu<1$ such than$$\lim_{x -> 0} x^{\mu+n}{{\log x}\over {1+x}}$$ exists
The limit converges for $\mu+n>0$ or $\mu>-n$ This is true for all $n>0$ Hence the given integral must converge for all $n>0$
The answer however given is that the integral is convergent for $n>-1$, How do i deduce that, ans is there another simpler work around ?
If $n>0$ the integrand is bounded at zero, as you can see using L'Hospital rule.
If $-1<\mu<n<0$, then
$$\dfrac{|x^n\ln x|}{1+x^2} < x^\mu$$
for $x$ sufficiently close to 0. Since the integral of $x^\mu$ converges, your integral converges too.