integral from 0 to 1 of 1/((2x-1)^2)

解答

\int_{0}^{1} \frac{1}{( 2 x - 1 ) ^{2}} d x

发散

求解步骤

\int_{0}^{1} \frac{1}{( 2 x - 1 ) ^{2}} d x

找到无定义的点（奇点）:

x = \frac{1}{2}

如果存在

b, a < b < c, f (b) =

无定义

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{\frac{1}{2}} \frac{1}{( 2 x - 1 ) ^{2}} d x + \int_{\frac{1}{2}}^{1} \frac{1}{( 2 x - 1 ) ^{2}} d x

\int_{0}^{\frac{1}{2}} \frac{1}{( 2 x - 1 ) ^{2}} d x =

发散

如果表达式有部分发散，则整个表达式发散

= 发散

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