Inspired by Kishore S Shenoy

We start with a bit of variable manipulation: $\begin{array}{rcl} \displaystyle I & = &\displaystyle \int_0^1 \frac{\ln(1 + x - x^2)}{x}\,dx \; = \; \left(\int_0^{\frac12} + \int_{\frac12}^1\right)\,\frac{\ln(1 + x(1-x))}{x}\,dx \\ & = & \displaystyle \int_0^{\frac12} \frac{\ln(1 + x(1-x))}{x}\,dx + \int_0^{\frac12}\frac{\ln(1 + x(1-x))}{1-x}\,dx \\ & = & \displaystyle\int_0^{\frac12} \frac{\ln(1 + x(1-x))}{x(1-x)}\,dx \\ & = & \displaystyle \int_0^{\frac12} \left(\sum_{m=0}^\infty \frac{(-1)^m}{m+1} x^m(1-x)^m\right)\,dx \; = \; \sum_{m=0}^\infty \frac{(-1)^m}{m+1}\int_0^{\frac12} x^m(1-x)^m\,dx \\ & = & \displaystyle\sum_{m=0}^\infty \frac{(-1)^m}{m+1} \int_0^{\frac14\pi} \sin^{2m}\theta \cos^{2m}\theta\,2\sin\theta\cos\theta\,d\theta \\ & = &\displaystyle 2\sum_{m=0}^\infty \frac{(-1)^m}{m+1}\int_0^{\frac14\pi} \sin^{2m+1}\theta \cos^{2m+1}\theta\,d\theta \; = \; 2\sum_{m=0}^\infty \frac{(-1)^m}{(m+1) 2^{2m+1}}\int_0^{\frac14\pi} \sin^{2m+1}2\theta\,d\theta \\ & = & \displaystyle \sum_{m=0}^\infty \frac{(-1)^m}{(m+1) 2^{2m+1}} \int_0^{\frac12\pi} \sin^{2m+1}\theta\,d\theta \; = \; \sum_{m=0}^\infty \frac{(-1)^m}{(m+1)2^{2m+1}} \frac{(2m)!!}{(2m+1)!!} \end{array}$ using the substitution $x = \sin^2\theta$ . Looking this final series up in Gradshteyn and Ryzhik (well, G&R give the formula for $(\sin^{-1}x)^2$ , but close enough) gives $I \; = \; 2\Big[\sinh^{-1}\tfrac12\Big]^2 \; = \; 2\Big[\ln\big(\tfrac12(1 + \sqrt{5})\big)\Big]^2 \;,$ so that $A \,=\, \tfrac12(1 + \sqrt{5})$ . Since $A^2 \,=\, A + 1$ , we have $A^3 \,=\, A^2 + A \,=\, 2A + 1$ , so that $A^3 - 3 \,=\, 2(A-1)$ . Thus $A(A^3-2) \,=\, 2A(A-1) \,=\, \boxed{2}$ .

I am sure that there must be a cute substitution that gets all the way to the last step, without the need for infinite series.

$I=\int _{ 0 }^{ 1 }{ \frac { ln(1+x-{ x }^{ 2 }) }{ x } dx } \\ Roots\quad of\quad (1+x-{ x }^{ 2 })\quad are\\ x=\frac { -1\pm \sqrt { 5 } }{ -2 } \\ let\quad \\ a=\frac { -1+\sqrt { 5 } }{ 2 } \\ b=\frac { -1-\sqrt { 5 } }{ -2 } =\frac { 1+\sqrt { 5 } }{ 2 } \\ I=\int _{ 0 }^{ 1 }{ \frac { ln[(x+a)(b-x)] }{ x } dx } \\ I=\int _{ 0 }^{ 1 }{ \frac { ln(1+\frac { x }{ a } )+ln(1-\frac { x }{ b } ) }{ x } dx } \\ (ln(ab)=0\quad since\quad ab=1)\\ I=\int _{ 0 }^{ 1 }{ \frac { ln(1+\frac { x }{ a } ) }{ x } +\quad \frac { ln(1-\frac { x }{ b } ) }{ x } dx } \\ Using\quad taylors\quad expansion\\ I=\int _{ 0 }^{ 1 }{ \frac { 1 }{ a } \sum _{ 1 }^{ \infty }{ \frac { { { (-1) }^{ n-1 }\left( \frac { x }{ a } \right) }^{ n-1 } }{ { n }^{ 2 } } } } dx-\int _{ 0 }^{ 1 }{ \frac { 1 }{ b } \sum _{ 1 }^{ \infty }{ \frac { { \left( \frac { x }{ b } \right) }^{ n-1 } }{ { n }^{ 2 } } } } dx\\ I=-\sum _{ 1 }^{ \infty }{ \frac { { \left( -\frac { 1 }{ a } \right) }^{ n } }{ { n }^{ 2 } } } -\sum _{ 1 }^{ \infty }{ \frac { { \left( \frac { 1 }{ b } \right) }^{ n } }{ { n }^{ 2 } } } \\ I=-{ Li }_{ 2 }\left( -\frac { 1 }{ a } \right) -{ Li }_{ 2 }\left( \frac { 1 }{ b } \right) \quad \quad [Li=Polylogarithmic\quad function]\\ I\simeq 0.463130\\ 2\left( ln(A) \right) ^{ 2 }=0.46130\\ A=0.618034\quad or\quad 1.618034\\ If\quad A=0.618034\\ Ans=-1.708195\\ If\quad A=1.618034\\ Ans=2\\ [Can't\quad find\quad a\quad proper\quad reason\quad why\quad second\quad one\quad is\quad right\quad \\ and\quad not\quad the\quad first\quad one,\quad \\ comments\quad appreciated]\\ Ans=\boxed { 2 }$