"j(t)=\\dfrac{1}{t^2-2t+1}"
"j(t)=\\dfrac{1}{(t-1)^2}" "t\\not=1"
Doman: "(-\\infin, 1)\\cup(1, \\infin)."
If we consider the interval "[-5, 5]," then "t\\in[-5, 1)\\cup(1, 5]."
"\\def\\arraystretch{1.5}\n \\begin{array}{c:c}\n t & y=j(t) \\\\ \\hline\n -5 & 1\/36 \\\\\n \\hdashline\n -4.5 & 4\/121 \\\\\n \\hdashline\n -4 & 1\/25 \\\\\n \\hdashline\n -3.5 & 4\/81 \\\\\n \\hdashline\n -3 & 1\/16 \\\\\n \\hdashline\n -2.5 & 4\/49 \\\\\n \\hdashline\n -2 & 1\/9 \\\\\n \\hdashline\n -1.5 & 4\/25 \\\\\n \\hdashline\n -1 & 1\/4 \\\\\n \\hdashline\n -0.5 & 4\/9 \\\\\n \\hdashline\n0 & 1 \\\\\n \\hdashline\n 0.5 & 4 \\\\\n \\hdashline\n 0.6 & 25\/4 \\\\\n \\hdashline\n 0.7 & 100\/9 \\\\\n \\hdashline\n 0.8 & 25 \\\\\n \\hdashline\n 0.9 & 100 \\\\\n \\hdashline\n 1.1 & 100 \\\\\n \\hdashline\n 1.2 & 25\\\\\n \\hdashline\n 1.3 & 100\/9 \\\\\n \\hdashline\n 1.4 & 25\/4 \\\\\n \\hdashline\n 1.5 & 4 \\\\\n \\hdashline\n 2 & 1 \\\\\n \\hdashline\n2.5 & 4\/9 \\\\\n \\hdashline\n3 & 1\/4 \\\\\n \\hdashline\n 3.5 & 4\/25 \\\\\n \\hdashline\n 4 & 1\/9 \\\\\n \\hdashline\n4.5 & 4\/49 \\\\\n \\hdashline\n 5 & 1\/16 \\\\\n\n\\end{array}"
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