How to solve this definite integral? It cannot be solved using simple integration by parts.

Using integration by parts, we can get $$\begin{align} \int_0^1{\frac{x}{x^2+1}\ln \left( 1+x \right) \mathrm{d}x}=&\frac{1}{2}\int_0^1{\frac{1}{x^2+1}\ln \left( 1+x \right) \mathrm{d}\left( x^2+1 \right)} \\ =&\frac{1}{2}\int_0^1{\ln \left( 1+x \right) \mathrm{d}\ln \left( x^2+1 \right)} \\ =&\frac{1}{2}\left( \ln ^22-\int_0^1{\frac{\ln \left( 1+x^2 \right)}{1+x}\mathrm{d}x} \right) \end{align}$$ And the second integral $\int_0^1{\frac{\ln \left( 1+x^2 \right)}{1+x}\mathrm{d}x}$ can be found in this cite, for example, from here we can get that $$ \int_{0}^{1}\frac{\ln(x^{2}+1)}{x+1}\mathrm{d}x=\frac{3}{4}\ln^{2}(2)-\frac{{\pi}^{2}}{48} $$ So combining them, we can get $$\begin{align} \int_0^1{\frac{x}{x^2+1}\ln \left( 1+x \right) \mathrm{d}x}=&\frac{1}{2}\left( \ln ^22-\left( \frac{3}{4}\ln ^22-\frac{\pi ^2}{48} \right) \right) \\ =&\frac{\pi ^2}{96}+\frac{1}{8}\ln ^22 \end{align}$$