#### Answer

$$T_4= (x-1)+\frac{1}{2}(x-1)^{2}-\frac{1}{6}(x-1)^{3}+\frac{1}{12}(x-1)^{4}$$

#### Work Step by Step

Given $$f(x)=x \ln (x), \quad T_{4}, \quad a=1$$
Since
\begin{align*}
f(x)&=x \ln (x)\ \ \ \ \ \ \ &f(1)=0\\
f'(x)&= 1+ \ln (x)\ \ \ \ \ \ \ &f(1)=1\\
f''(x)&=\frac{1}{x}\ \ \ \ \ \ \ &f(1)=1\\
f'''(x)&=\frac{-1}{x^2}\ \ \ \ \ \ \ &f(1)=-1\\
f^{(4)}(x)&=\frac{2}{x^3}\ \ \ \ \ \ \ &f(1)=2
\end{align*}
Then
\begin{aligned}
T_{4}(x)&= f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3}++\frac{f^{(4)}(a)}{4 !}(x-a)^{4} \\
&= (x-1)+\frac{1}{2 !}(x-1)^{2}-\frac{1}{3 !}(x-1)^{3}+\frac{2}{4 !}(x-1)^{4}\\
&=(x-1)+\frac{1}{2}(x-1)^{2}-\frac{1}{6}(x-1)^{3}+\frac{1}{12}(x-1)^{4}
\end{aligned}