Q. $\int_{}^{}\frac{3 - x^{2}}{1 - 2x + x^{2}}.e^{x}dx = e^{x}f(x) + c \Rightarrow f(x)$
A) $\frac{1 + x}{1 - x}$
B) $\frac{1 - x}{1 + x}$
C) $\frac{1 + x}{x - 1}$
D) $\frac{x - 1}{1 + x}$
Explanation
(a)
Let $I = \int_{}^{}\frac{3 - x^{2}}{(1 - x)^{2}}.e^{x}dx = \int_{}^{}{\left( \frac{2}{(1 - x)^{2}} + \frac{1 + x}{1 - x} \right)e^{x}dx}$
$\Rightarrow \ I = e^{x}\left( \frac{1 + x}{1 - x} \right) + c$
But $I = e^{x}f(x)$+c [given]
$\therefore\ f(x) = \frac{1 + x}{1 - x}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.