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In order to compute, in an elementary way,

$displaystyle int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$

(see Evaluating $int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$ )

i need to show, in a simple way, that:

$displaystyle int_0^{tfrac{pi}{4}}ln(cos x-sin x)ln(cos x) dx - int_0^{tfrac{pi}{4}}ln(cos x+sin x)ln(sin x)dx=dfrac{Gln 2}{2}$

$G$ being the Catalan constant.

The reason of my interest for this question is, if i am right, that this formula permits to find out a relation between integrals:

$displaystyle int_0^1 dfrac{ln(1+x)ln(1+x^2)}{1+x^2}dx$

$displaystyle int_0^1 dfrac{ln(1+x)^2}{1+x^2}dx$

$displaystyle int_0^{tfrac{pi}{4}}big(ln(cos x)big)^2 dfrac{}{}dx$

and some constants.

Using Simpsons rule we know
$$cos(x)-sin(x)=sqrt{2}sinleft(frac{pi}{4}-xright)$$ and
$$cos(x)+sin(x)=sqrt{2}cosleft(frac{pi}{4}-xright)$$

Plugging this in the original integrals yields:
$$int_0^{frac{pi}{4}}lnleft(sqrt{2}sinleft(frac{pi}{4}-xright)right)lnleft(cos(x)right)mathrm{d}x-
int_0^{frac{pi}{4}}lnleft(sqrt{2}cosleft(frac{pi}{4}-xright)right)lnleft(sin(x)right)mathrm{d}x$$

Splitting the logarithms and rearranging yields:
$$int_0^{frac{pi}{4}}ln(sqrt{2})left(ln(cos(x))-ln(sin(x)right)mathrm{d}x+\
+left(int_0^{frac{pi}{4}}lnleft(sinleft(frac{pi}{4}-x right)right)ln(cos(x)) mathrm{d}x-int_0^{frac{pi}{4}}lnleft(cosleft(frac{pi}{4}-x right)right)ln(sin(x)) mathrm{d}x right)$$

The first part yields
$$int_0^{frac{pi}{4}}ln(sqrt{2})left(ln(cos(x))-ln(sin(x)right)mathrm{d}x = ln(sqrt{2})G=frac{Gln(2)}{2}$$

and because $$int_0^bf(x)mathrm{d}x=int_0^bf(b-x)mathrm{d}x$$
the second part yields
$$int_0^{frac{pi}{4}}lnleft(sinleft(frac{pi}{4}-x right)right)ln(cos(x)) mathrm{d}x-int_0^{frac{pi}{4}}lnleft(cosleft(frac{pi}{4}-x right)right)ln(sin(x)) mathrm{d}x=0$$

begin{align}
I&:=int_0^{tfrac{pi}{4}}ln(cos x-sin x)ln(cos x),dx - int_0^{tfrac{pi}{4}}ln(cos x+sin x)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},cosleft(x+frac {pi}4right)right)ln(cos x),dx - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
& quadtext{setting}; x=frac {pi}4-y; text{in the first integral gives}\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},cosleft(frac {pi}2-yright)right)lnleft(cosleft(frac {pi}2-frac {pi}4-yright)right),dy - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(yright)right)lnleft(sinleft(y+frac {pi}4right)right),dy - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}left(ln(sqrt{2})+lnleft(sin xright)right)lnleft(sinleft(x+frac {pi}4right)right) - left(ln(sqrt{2})+lnleft(sinleft(x+frac {pi}4right)right)right)ln(sin x);dx\
&=ln(sqrt{2})int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x);dx\
&=dfrac{ln 2}{2}G\
end{align}
It remains to prove that $displaystyle int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x);dx=G$.

This is detailed around $(16)$ in this interesting paper by Jameson and Lord or using :
begin{align}
int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x),dx&=int_{tfrac{pi}{4}}^{tfrac{pi}{2}}lnleft(sin xright),dx-int_0^{tfrac{pi}{4}}ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}ln(cos x),dx-int_0^{tfrac{pi}{4}}ln(sin x),dx\
&=-int_0^{tfrac{pi}{4}}ln(tan x),dx\
&=-int_0^1frac{ln t}{1+t^2},dt,quadtext{integrated by parts}\
&=-left.ln(t);arctan(t)right|_0^1+int_0^1frac{arctan(t)}{t},dt\
&=int_0^1sum_{n=0}^infty (-1)^nfrac{t^{2n}}{2n+1},dt\
&=left.sum_{n=0}^infty (-1)^nfrac{t^{2n+1}}{(2n+1)^2}right|_0^1\
&=G\
end{align}

Relationship between the proposed integrals.

$displaystyle int_0^1 dfrac{ln(1+x)^2}{1+x^2}dx$-$displaystyle int_0^1 dfrac{ln(1+x)ln(1+x^2)}{1+x^2}dx=-frac{3pi}{16}ln^2{2}+dfrac{Gln 2}{2}$