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

Question

Aicharya · Accepted Answer

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} $

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

Solution