I am not even sure if I can apply contour integration over this. But I tired and saw that $(z+1)^2$ gets cancelled in the denominator giving me no real part and only $-2i \pi$.
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In this answer it's shown using contour integration that $$ \int_0^{\infty} \frac{\ln^2(x)}{(1+x)^2}\, dx = \frac{\pi^2}{3} $$ Now just notice that \begin{align} \int_0^{\infty} \frac{\ln^2(x)}{(1+x)^2}\, dx &=\int_0^{1} \frac{\ln^2(x)}{(1+x)^2}\, dx + \int_{1}^{\infty} \frac{\ln^2(x)}{(1+x)^2}\, dx\\ & \overset{\color{blue}{u = 1/x}}{=}\int_0^{1} \frac{\ln^2(x)}{(1+x)^2}\, dx + \int_{1}^{0}\frac{\ln^2\left(\frac{1}{u} \right)}{\left(1 + \frac{1}{u^2} \right)} \left( -\frac{1}{u^2}\right) \, du\\ & = 2\int_0^{1} \frac{\ln^2(x)}{(1+x)^2}\, dx \end{align} So you can conclude that $$ \boxed{\int_0^{1} \frac{\log^2{x}}{(1+x)^2}\, dx = \frac{\pi^2}{6}} $$
Also, in the same question linked one of the other answers gives a non-contour integration way to get the result using Taylor series. So you can also use that route if you like.
Robert Lee
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$\int_{0}^1 \ln^2 (x) / (x+1)^2 dx$– Jean Marie Jun 19 '21 at 08:47