已知函数f(x)={(x−1)arctan1x−1+2,x<1a,x=1x+b,x>1f(x)=\begin{cases}(x - 1)\arctan\dfrac{1}{x - 1}+2, & x < 1\\a, & x = 1\\x + b, & x > 1\end{cases}f(x)=⎩⎨⎧(x−1)arctanx−11+2,a,x+b,x<1x=1x>1在点x=1x = 1x=1处连续，求a,ba,ba,b的值.
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