All Roads Lead To $\pi$

We can write the above Integral as an integral of an infinite series,like so:

$\int _{ 0 }^{ 1 }{ ({ \ln { x) } }^{ 2 }{ (1-x) }^{ -2 } } dx=\int _{ 0 }^{ 1 }{ { (\ln { x) } }^{ 2 }(1+2x+3{ x }^{ 2 }+4{ x }^{ 3 }......) } dx$

A general form would be ,

$I=\sum _{ r=0 }^{ \infty }{ \int _{ 0 }^{ 1 }{ { (lnx) }^{ 2 } } } (r+1){ x }^{ r }dx$

now $I(m,n)=\int _{ 0 }^{ 1 }{ { x }^{ m } } { (lnx) }^{ n }dx=\frac { { (-1 })^{ n }n! }{ { (m+1) }^{ n+1 } }$

The derivation of the above expression is part of the solution of an amazing and very similar problem by Ishan Tarunesh,here's the link: Off the charts!

Therefore, On solving further

$\sum _{ r=0 }^{ \infty }{ (r+1)\frac { { (-1) }^{ 2 }2! }{ { (r+1) }^{ 2+1 } } } \quad \quad =\quad \quad 2\sum _{ r=0 }^{ \infty }{ \frac { 1 }{ { (r+1 })^{ 2 } } \quad \quad = } \frac { { \pi }^{ 2 } }{ 3 }$

And so $a=\boxed{2}$