integral from 0 to 4 of 1/(x^2-x-2)

解答

\int_{0}^{4} \frac{1}{x ^{2} - x - 2} d x

发散

求解步骤

\int_{0}^{4} \frac{1}{x ^{2} - x - 2} d x

找到无定义的点（奇点）:

x = 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}^{2} \frac{1}{x ^{2} - x - 2} d x + \int_{2}^{4} \frac{1}{x ^{2} - x - 2} d x

\int_{0}^{2} \frac{1}{x ^{2} - x - 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

\int_{0}^{π} x^{2} cos (n x) d x

\frac{d}{d x} (x^{3} + 3 x^{2} - 24 x - 80)

\frac{d}{d x} (\frac{3 sin ( x )}{3 + cos ( x )})

im pl i c i t \frac{d y}{d x}, e^{x y} + x^{2} - y^{2} = 12

n = 0 \sum \infty (\frac{1}{17})^{n}