How do you proceed for the following? I know I have to treat y as a constant to integrate but I am stuck on how to do it. I tried by parts without success I am sure there is a trick to it?
$\underset{y \mapsto 1 ^{-}}{lim}\int_{ln{y}}^{0} (\frac{xln(e^{x}-x+x^{2})}{(y-1)^{3}})dx$
$$L=\lim_{y\to 1^-}\int_{\ln{y}}^{0}\frac{x\ln(e^{x}-x+x^{2})}{(y-1)^{3}}dx=$$
$$=\lim_{y\to 1^-}\dfrac{\int_{\ln{y}}^{0}x\ln(e^{x}-x+x^{2})}{(y-1)^{3}}dx$$
Being a limit of the type $0/0$ (as the integral vanishes if $y=1$ because $\ln y=0$), we can apply the l'Hôpital's rule. For the derivative of the integral, we have to use the Leibniz rule:
$$L=\lim_{y\to 1^-}\dfrac{-\ln y\ln(y-\ln y+\ln^2y)/y}{3(y-1)^2}$$
Again a $0/0$ type. It's easy to follow from here using l'Hôpital again, maybe twice.