$$ \int_0^1 \dfrac{\ln^3 \big(\frac{1-x}{1+x} \big)}{x} dx $$
Put $u = \frac{1-x}{1+x}$
Using componendo-dividendo, I get,
$x= \frac{1-u}{1+u} \implies du = -\frac{2}{(1+u)^2} $
$$ \int_0^1 -\dfrac{2 \ln^3(u)}{(1-u)(1+u)}du $$
Write $2$ as $ \big[ (1+u) + (1 - u) \big] $ and seperate the common denominator.
$$ = - \Bigg[ \int_0^1 \dfrac{\ln^3(u)}{1-u} du + \int_0^1 \dfrac{ \ln^3 (u)}{1+u} du \Bigg] $$
Okay I know the values of those 2 integral, so I'll directly use it.
$$= - \Bigg[ - \dfrac{π^4}{15} - \dfrac{7π^4}{120} \Bigg] $$
$$= - \Bigg( - \dfrac{π^4}{8} \Bigg)$$
$$ = \dfrac{π^4}{8}$$
But when I put it into the calculator, the answer is :
$- \dfrac{π^4}{8} $
Where did I miss the negative sign?