integral from 0 to P of 1/(M-P)

解答

\int_{0}^{P} \frac{1}{M - P} d P

- ln ∣ M - P ∣ + ln ∣ M ∣

求解步骤

\int_{0}^{P} \frac{1}{M - P} d P

使用换元积分法

= \int_{M}^{M - P} - \frac{1}{u} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int_{M}^{M - P} \frac{1}{u} d u

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - [ln ∣ u ∣]_{M}^{M - P}

带入上下限计算后得:

ln ∣ M - P ∣ - ln ∣ M ∣

= - (ln ∣ M - P ∣ - ln ∣ M ∣)

化简

= - ln ∣ M - P ∣ + ln ∣ M ∣

s l o p e in t erce pt (1) (0.3)

\int_{- 1}^{1} x^{3} (1 + x^{4})^{3} d x

t an g e n t f (x) = 8 + 5 x^{2} - 2 x^{3}, (1, 11)

t an g e n t f (x) = \frac{4}{x}, (4, 1)

\int_{0}^{l n (3)} 2 (3 - x e^{x}) d x