$\int \left(\frac{{\displaystyle 3-x^2}}{{\displaystyle 1-2x+x^2}}\right)e^{x}dx=e^x.f\left(x\right)+c\Rightarrow f\left(x\right)=$

$\int \left(\frac{{\displaystyle 3-x^2}}{{\displaystyle 1-2x+x^2}}\right)e^{x}dx$

$=\int e^x\left[\frac{2+(1-x^2)}{{(1-x)}^2}\right]dx$

$=\int e^x\left[\frac{2}{{(1-x)}^2}+\frac{(1+x)(1-x)}{{(1-x)}^2}\right]dx$

$=\int e^x\left[\frac{1+x}{1-x}+\frac{2}{{(1-x)}^2}\right]dx$

$$\begin{gathered} =\int e^x[f\left(x\right)+f\text{'}\left(x\right)]dx \\ wheref\left(x\right)=\frac{1+x}{1-x}\Rightarrow f\text{'}\left(x\right)=\frac{(1-x)(1+x)}{{(1-x)}^2} \end{gathered}$$

$=e^{x}f\left(x\right)+c$

$=e^x\left(\frac{{\displaystyle 1+x}}{{\displaystyle 1-x}}\right)+c$