${\int }_{-1}^1\frac{\log (1+x)}{1+x^2}dx={\int }_0^1\frac{\log (1+x)}{1+x^2}dx+{\int }_0^{1}f\left(x\right)dx$ then $f\left(x\right)=$

Given ${\int }_{-1}^1\frac{log(1+x)}{1+x^2}dx={\int }_{-1}^0\frac{log(1+x)}{1+x^2}dx +{\int }_0^1\frac{log(1+x)}{1+x^2}dx$

$\left[\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(x\right)dx+{\int }_b^{c}f\left(x\right)dx\right]$

Now, 

Let $I_1={\int }_{-1}^0\frac{log\left(1+x\right)}{1+x^2}dx$

Put $x=-t\Rightarrow dx=-dt$

when $x=-1, t=1 \& when x=0, t=0$

$$\begin{gathered} \therefore I_1={\int }_1^0\frac{log\left(1-t\right)}{1+t^2}\left(-dt\right) \\ \Rightarrow I_1={\int }_0^1\frac{log\left(1-t\right)}{1+t^2}dt \left[\because {\int }_a^{b}f\left(x\right)dx=-{\int }_b^{a}f\left(x\right)dx\right] \end{gathered}$$

$\therefore I_1={\int }_0^1\frac{log\left(1-x\right)}{1+x^2}dx \left[\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(t\right)dt\right]$

$\therefore {\int }_{-1}^1\frac{log\left(1+x\right)}{1+x^2}dx={\int }_0^1\frac{log\left(1+x\right)}{1+x^2}dx+{\int }_0^{1}f\left(x\right)dx$

$={\int }_0^1\frac{log\left(1+x\right)}{1+x^2}dx+{\int }_0^1\frac{log\left(1-x\right)}{1+x^2}dx$

$\therefore \mathrm{f}\left(\mathrm{x}\right)=\frac{\log \left(1-\mathrm{x}\right)}{1+{\mathrm{x}}^2}$