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

解答

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

- ln ∣ - x + 1 ∣

求解步骤

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

使用换元积分法

= \int_{1}^{1 - x} - \frac{1}{u} d u

提出常数:

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

= - \int_{1}^{1 - x} \frac{1}{u} d u

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - [ln ∣ u ∣]_{1}^{1 - x}

带入上下限计算后得:

ln ∣ - x + 1 ∣

= - ln ∣ - x + 1 ∣

x \to 0 lim (\frac{1}{1 - e ^{x}})

n = 1 \sum \infty (cos (32))^{n}

\frac{\partial}{\partial x} (x^{3} - 9 x y + 7 y^{3})

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ma c l a u r in cos (x^{4})