Question

If $\left|x\right|<1$, the coefficient of $x^2$ in the expansion of $\frac{{\log }_e(1+x)}{{(1-x)}^2}$

• A. $\frac{3}{2}$
• B. ${}^2{C}_1$
• C. $-\frac{1}{2}$
• D. none of these

Answer 1

The correct option is A $\frac{3}{2}$

Given, $\log (1+x)(1-x{)}^{-2}$
$=(x-\frac{x^2}{2}+\frac{x^3}{3}...\infty )(1+2x+3x^2+...\infty )$
Hence term with $x^2$ will be
$x.\left(2x\right)-\frac{x^2}{2}.1$
$=\frac{3}{2}{x}^2$
Hence coefficient is
$=\frac{3}{2}$