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For start we'll prove a result that it's going to be used later. $$\boxed{\int_0^1 \frac{\ln(1+x)}{x}dx=\frac12 \int_0^1 \frac{\ln x}{x-1}dx}=\frac12\cdot \frac{\pi^2}{6}$$ proof: $$\int_0^1 \frac{\ln x}{x+1}dx+\int_0^1 \frac{\ln x}{x-1}dx=\int_0^1 \frac{2x\ln x}{x^2-1}dx\overset{x^2\to x}=\frac12 \int_0^1 \frac{\ln x}{x-1}dx $$ $$\Rightarrow \int_0^1 \frac{\ln x}{x+1}dx=-\frac12 \int_0^1 \frac{\ln x}{x-1}dx$$ $$\int_0^1 \frac{\ln(1+x)}{x}dx=\underbrace{\ln x \ln(1+x)\bigg|_0^1}_{=0}-\int_0^1 \frac{\ln x}{1+x}dx=\frac12 \int_0^1 \frac{\ln x}{x-1}dx$$

Now back to the question. Consider the following integrals:$$I=\int_0^1\frac{x\ln (1+x)}{1+x^2}dx,\quad J=\int_0^1\frac{x\ln (1-x)}{1+x^2}dx$$

$$I+J=\int_0^1 \frac{x\ln(1-x^2)}{1+x^2}dx\overset{x^2=t}=\frac12\int_0^1 \frac{\ln(1-t)}{1+t}dt$$ Now we will integral by parts, however we can't chose $\ln(1+t)'=\frac{1}{1+t}$ since we run into divergence issues.

We will take $(\ln (1+t)-\ln 2)'=\frac{1}{1+t}$ then: $$I+J=\frac12\bigg[\underbrace{\ln(1-t)(\ln(1+t)-\ln 2)\bigg]_0^1}_{=0}+\frac12 \int_0^1 \frac{\ln\left(\frac{1+t}{2}\right)}{1-t}dt$$ Now substitute $t=\frac{1-y}{1+y}$ in order to get: $$I+J=-\frac12 \int_0^1 \frac{\ln(1+y)}{y(1+y)}dy=\frac12 \int_0^1 \frac{\ln(1+y)}{1+y}dy-\frac12\int_0^1 \frac{\ln(1+y)}{y}dy$$ $$=\frac14 \ln^2(1+y)\bigg|_0^1 -\frac14 \cdot \frac{\pi^2}{6}=\boxed{\frac{\ln^2 2}{4}-\frac{\pi^2}{24}}$$

Similarly, for $I-J$ set $\frac{1-x}{1+x}= t$ to get: $$I-J=-\int_0^1 \frac{x\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}dx=\underbrace{\int_0^1 \frac{t\ln t}{1+t^2}dt}_{t^2\rightarrow t}-\int_0^1 \frac{\ln t}{1+t}dt$$ $$=\frac14 \int_0^1 \frac{\ln t}{1+t}dt-\int_0^1 \frac{\ln t}{1+t}dt=-\frac34 \left(\underbrace{\ln(1+t)\ln t \bigg|_0^1}_{=0} -\int_0^1 \frac{\ln(1+t)}{t}dt\right)$$ $$=\frac34 \int_0^1 \frac{\ln(1+t)}{t}dt=\frac38 \int_0^1 \frac{\ln t}{t-1}dt=\boxed{\frac{\pi^2}{16}}$$

Finally we can extract the integral as: $$I=\frac12 \left((I+J)+(I-J)\right)=\frac12\left(\frac{\ln^2 2}{4}-\frac{\pi^2}{24}+\frac{\pi^2}{16}\right)=\boxed{\frac{\ln^2 2}{8}+\frac{\pi^2}{96}}$$ Supplementary result is following: $$\boxed{\int_0^1\frac{x\ln (1-x)}{1+x^2}dx=\frac{\ln^2 2}{8}-\frac{5\pi^2}{96}}$$