$\int \frac{3-x^2}{1-2 x+x^2} e^x d x=e^x f(x)+c \Rightarrow f(x)$

• Correct answer
  A

  $\frac{1+x}{1-x}$
• B

  $\frac{1-x}{1+x}$
• C

  $\frac{1+x}{x-1}$
• D

  $\frac{x-1}{1+x}$

Solution

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 $$ \begin{aligned} & e^x\left(\frac{1+x}{1-x}\right)+c=e^x f(x)+c \\ & \Rightarrow \quad f(x)=\frac{1+x}{1-x} \\ & \end{aligned} $$

Hence, the correct answer is option: 'A'.