Answer
$x-\dfrac{x^2}{2}+\dfrac{x^3}{3}$
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: $f(x)=\ln(1+x) \implies f(0)=0$ with $n=3$ and $a=0$
Further, $f'(x)=\dfrac{1}{1+x} \implies f'(0)=1 \\ f''(x)=\dfrac{-1}{(x+1)^2} \implies f''(0)=-1$ and $f'''(x)=\dfrac{2}{(x+1)^3} \implies f'''(0)=2$
Now, plug these values in the equation (1) to obtain:
$P_3(x)=f(0)+f'(0)\dfrac{x}{1!}+\dfrac{1}{2!}f''(0)x^2+\dfrac{1}{3!}f'''(0)x^3\\=0+x- \dfrac{x^2}{2!}+\dfrac{2x^3}{3!}\\=x-\dfrac{x^2}{2}+\dfrac{2x^3}{6}\\=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}$
