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

解答

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

发散

求解步骤

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

提出常数:

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

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

使用换元积分法

= 2 \cdot \int_{0}^{- 4} \frac{1}{2 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 2 (- \int_{- 4}^{0} \frac{1}{2 u} d u)

提出常数:

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

= 2 (- \frac{1}{2} \cdot \int_{- 4}^{0} \frac{1}{u} d u)

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{2} [ln ∣ u ∣]_{- 4}^{0})

化简

2 (- \frac{1}{2} [ln ∣ u ∣]_{- 4}^{0}) : - [ln ∣ u ∣]_{- 4}^{0}

= - [ln ∣ u ∣]_{- 4}^{0}

带入上下限计算后得:

发散

= 发散

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

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

\int_{1}^{e} (19 + ln (x))^{2} d x