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$\int \frac{3-x^2}{1-2 x+x^2} e^x d x=e^x f(x)+c \Rightarrow f(x)$
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Verified Answer
The correct answer is:
$\frac{1+x}{1-x}$
Let $\begin{aligned} I & =\int \frac{3-x^2}{1-2 x+x^2} e^x d x \\ & =\int \frac{3-x^2}{(1-x)^2} e^x d x \\ & =\int\left(\frac{2}{(1-x)^2}+\frac{1+x}{1-x}\right) e^x d x\end{aligned}$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
& e^x\left(\frac{1+x}{1-x}\right)+c=e^x f(x)+c \\
& \Rightarrow \quad f(x)=\frac{1+x}{1-x} \\
&
\end{aligned}
$$
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