Answer
$A_{0}(x)=0 \\ A_{1}(x)=x \\A_{2}(x)=x-(\dfrac{1}{2})x^2 \\A_{3}(x)=x-(\dfrac{1}{2})x^2 +(\dfrac{1}{3})x^3$
Work Step by Step
The Taylor polynomial is defined for the function $f(x)$ at the point $a$ of order $n$ as:
$A_n(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+....+\dfrac{f^{n}(a)}{n!}(x-a)^n$
Now, we get $f(0)=0 \\ f'(x)=(1+x)^{-1} \implies f'(0)=1\\ f''(x)=-(1+x)^{-2} \implies f''(0)=-1 \\ f'''(x)=2(1+x)^{-3}\implies f'''(0)=2$
This implies that $A_{0}(x)=0 \\ A_{1}(x)=x \\A_{2}(x)=x-(\dfrac{1}{2})x^2 \\A_{3}(x)=x-(\dfrac{1}{2})x^2+(\dfrac{2}{6})(x-0)^3 =x-(\dfrac{1}{2})x^2 +(\dfrac{1}{3})x^3$
