TITLE: Show that $\int^1_0 \frac{\tanh^{-1} x}{x} \ln [(1 + x)^3 (1 - x)] \, dx = 0$
QUESTION [9 upvotes]: I am looking for a nice slick way to show

$$\int^1_0 \frac{\tanh^{-1} x}{x} \ln [(1 + x)^3 (1 - x)] \, dx = 0.$$

So far I can only show the result using brute force as follows.
Let 
$$I = \int^1_0 \frac{\tanh^{-1} x}{x} \ln [(1 + x)^3 (1 - x)] \, dx.$$
Since
$$\tanh^{-1} x = \frac{1}{2} \ln \left (\frac{1 + x}{1 - x} \right ),$$
the above integral, after rearranging, can be rewritten as
$$I = \frac{3}{2} \int^1_0 \frac{\ln^2 (1 + x)}{x} \, dx - \int^1_0 \frac{\ln (1 - x) \ln (1 + x)}{x} \, dx - \frac{1}{2} \int^1_0 \frac{\ln^2 (1 - x)}{x} \, dx.\tag1$$
Each of the above three integrals can be found. The results are:
$$\int^1_0 \frac{\ln^2 (1 + x)}{x} \, dx = \frac{1}{4} \zeta (3).$$ For a proof, see here or here.
$$\int^1_0 \frac{\ln (1 - x) \ln (1 + x)}{x} \, dx = -\frac{5}{8} \zeta (3).$$
For a proof, see here. And
$$\int^1_0 \frac{\ln^2 (1 - x)}{x} \, dx = 2 \zeta (3).$$
For a proof of this last one, see here.
Thus (1) becomes
$$\int^1_0 \frac{\tanh^{-1} x}{x} \ln [(1 + x)^3 (1 - x)] \, dx = \frac{3}{8 } \zeta (3) + \frac{5}{8} \zeta (3) - \zeta (3) = 0,$$
as expected.

REPLY [10 votes]: $$\sf I = \frac12 \int^1_0 \frac{\ln\left(\frac{1+x}{1-x}\right)}{x} \ln ((1 + x)^3 (1 - x)) dx=\frac12 \int_0^1 \frac{(a-b)(3a+b)}{x}dx$$
Where we denoted $\sf a=\ln(1+x)$ and $\sf b=\ln(1-x)$.
Now we're going to use the following algebraic expression: $$\sf (a-b)(3a+b)=(a+b)^2+2(a-b)^2 -4b^2$$
Which is obtainable easily by combining the following expressions:
$$\sf a^2=\frac12(a+b)^2+\frac12(a-b)^2 -b^2,\quad ab=\frac14(a+b)^2-\frac14(a-b)^2\tag 1$$
$$\Rightarrow \sf 2I=\int_0^1  \frac{\ln^2\left(1-x^2\right)}{x}dx +2\int_0^1 \frac{\ln^2\left(\frac{1+x}{1-x}\right)}{x}dx-4\int_0^1 \frac{\ln^2\left(1-x\right)}{x}dx$$
Now we put $\sf x^2=t$ in the first part and $\sf \frac{1-x}{1+x}=t$ for the second one to get:
$$\sf 2I=\frac12 \int_0^1 \frac{\ln^2(1-t)}{t}dt+4\int_0^1 \frac{\ln^2 t }{1-t^2}dt-4\int_0^1 \frac{\ln^2(1-t)}{t}dt$$
$$\sf =-\frac72\int_0^1 \frac{\ln^2 t}{1-t}dt+4\int_0^1 \frac{\ln^2 t }{1-t^2}dt=-\frac72\int_0^1 \frac{\ln^2 t}{1-t}dt+\frac72\int_0^1 \frac{\ln^2 t}{1-t}dt=0$$

Above we used that:
$$\boxed{\sf \int_0^1 \frac{\ln^2 x}{1-x^2}dx=\frac78 \int_0^1 \frac{\ln^2 x}{1-x}dx}$$
And we can show this in two steps. First:
$$\sf {\int_0^1 \frac{\ln^2 x}{1-x}dx}\overset{x\to x^2}=8\int_0^1 \frac{x\ln^2 x}{1-x^2}dx=4{\int_0^1 \frac{\ln^2 x}{1-x}dx}-4\int_0^1 \frac{\ln^2 x}{1+x}dx$$
$$\sf \Rightarrow (1-4){\int_0^1 \frac{\ln^2 x}{1-x}dx}=-4\int_0^1 \frac{\ln^2 x}{1+x}dx\Rightarrow \boxed{\int_0^1 \frac{\ln^2 x}{1+x}dx=\frac34 \int_0^1 \frac{\ln^2 x}{1-x}dx}$$
But we also have that:
$$\sf \int_0^1 \frac{\ln^2 x}{1-x^2}dx=\frac12 \int_0^1 \frac{\ln^2 x}{1-x}dx+\frac12 \int_0^1 \frac{\ln^2 x}{1+x} dx $$
$$\sf =\frac12\int_0^1 \frac{\ln^2 x}{1-x}dx+ \frac38\int_0^1 \frac{\ln^2 x}{1-x}dx=\frac78 \int_0^1 \frac{\ln^2 x}{1-x}dx$$

Generalization. In a similar fashion we can deal with the following integral:
$$\sf I(m,n,q,p)=\int_0^1 \frac{[m\ln(1+x)+n\ln(1-x)][q\ln(1+x)+p\ln(1-x)]}{x}dx$$
Like from above we will keep $\sf a=\ln(1+x)$ and $\sf b=\ln(1-x)$. Thus we can write the numarator as:
$$\sf f=(ma+nb)(qa+pb)=mqa^2+(mp+nq)ab+npb^2$$
Using $(1)$ again we obtain:
$$\sf f=\left(\frac{mq}{2}+\frac{mp+nq}{4}\right)(a+b)^2+\left(\frac{mq}{2}-\frac{mp+nq}{4}\right)(a-b)^2+(np-mq)b^2$$
Furthermore we can write:
$$\sf \int_0^1 \frac{(a+b)^2}{x}dx=\int_0^1 \frac{\ln^2(1-x^2)}{x}dx=\frac12 \int_0^1 \frac{\ln^2 x}{1-x}dx$$
$$\sf \int_0^1 \frac{(a-b)^2}{x}dx=\int_0^1 \frac{\ln^2\left(\frac{1+x}{1-x}\right)}{x}dx=2\int_0^1 \frac{\ln^2 x}{1-x^2}dx=\frac74\int_0^1 \frac{\ln^2 x}{1-x}dx$$
$$\sf \Rightarrow I(m,n,q,p)=\left(\frac{mq}{8}-\frac{5}{16}(mp+nq)+np\right)\int_0^1 \frac{\ln^2 x}{1-x}dx$$
$$=\boxed{\sf \left(\frac{mq}{4}-\frac{5}{8}(mp+nq)+2np\right)\zeta(3)}$$