题目
$$\int_{5}^{+\infty} \frac{1}{x^{2}-4x+3} = ?$$
解析
因为：
$$
(x-2)^{2} = x^{2} + 4 – 4x.
$$
所以：
$$
x^{2} – 4x + 3 =
$$
$$
(x-2)^{2}-1 =
$$
$$
[(x-2)+1][(x-2)-1] =
$$
$$
(x-3)(x-1)
$$
于是：
$$
\int_{5}^{+\infty} \frac{1}{x^{2}-4x+3} =
$$
$$
\int_{5}^{+\infty} \frac{1}{(x-3)(x-1)}
$$
又：
$$
\frac{1}{x-3} – \frac{1}{x-1} =
$$
$$
\frac{2}{(x-3)(x-1)}
$$
所以：
$$
\int_{5}^{+\infty} \frac{1}{(x-3)(x-1)} =
$$
$$
\frac{1}{2} \int_{5}^{+ \infty}[ \frac{1}{x-3} – \frac{1}{x-1}] =
$$
$$
\frac{1}{2} [\ln |x-3| – \ln |x-1|] |_{5}^{+ \infty} =
$$
$$
\frac{1}{2} (\ln \frac{x-3}{x-1})|_{5}^{+ \infty} =
$$
$$
\frac{1}{2} (\ln \frac{+ \infty}{+ \infty} – \ln \frac{2}{4}) =
$$
$$
\frac{1}{2} (\ln 1 – \ln \frac{1}{2}) =
$$
$$
\frac{1}{2} (0 – \ln \frac{1}{2}) =
$$
$$
\frac{1}{2} \cdot (-1) \cdot \ln \frac{1}{2} =
$$
$$
\frac{1}{2} \ln (\frac{1}{2})^{-1} =
$$
$$
\frac{1}{2} \ln 2.
$$
综上可知，正确答案为 $\frac{1}{2} \ln 2$.
EOF