integral from-1 to 1 of (x-3+1/(x-2))

解答

\int_{- 1}^{1} (x - 3 + \frac{1}{x - 2}) d x

- ln (3) - 6

十进制

- 7.09861 \dots

求解步骤

\int_{- 1}^{1} x - 3 + \frac{1}{x - 2} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{1} x d x - \int_{- 1}^{1} 3 d x + \int_{- 1}^{1} \frac{1}{x - 2} d x

\int_{- 1}^{1} x d x = 0

\int_{- 1}^{1} 3 d x = 6

\int_{- 1}^{1} \frac{1}{x - 2} d x = - ln (3)

= 0 - 6 - ln (3)

化简

= - ln (3) - 6

\int_{- 3}^{2} (9 - x^{2}) - (x + 3) d x

\int_{- 1}^{3} 3 x^{2} - 6 x - 9 d x

\int_{0}^{5} 1 + \frac{9}{4} x d x

\int_{1}^{3} (3 x + 2) d x

\int_{0}^{2 π} - sin (t) d t