Answer
$x- 1-\dfrac{(x-1)^2}{2}$
Work Step by Step
The Taylor approximation for degree $n$ centered at point $a$ can be written as:
$P_n(x)=f(a)+f'(a)(x-a)+\dfrac{1}{2}f''(a)(x-a)^2+.......+\dfrac{1}{n!}f^n(a)(x-a)^n ~~~~........(1)$----
We have: $f(x)=\ln x \implies f(1)=0$ with $n=2$ and $a=1$
Further, $f'(x)=\dfrac{1}{x} \implies f'(1)=1 \\ f''(x)=\dfrac{-1}{x^2} \implies f''(1)=-1$
Now, plug these values in the equation (1) to obtain:
$P_2(x)=f(1)+f'(1)\dfrac{(x-1)}{1!}+\dfrac{1}{2!}f''(1)(x-1)^2\\=x- 1-\dfrac{(x-1)^2}{2}$
