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受欢迎的微积分 >

integral from 0 to 7 of 1/(u^2-49)

解答

\int_{0}^{7} \frac{1}{u ^{2} - 49} d u

解答

发散

求解步骤

\int_{0}^{7} \frac{1}{u ^{2} - 49} d u

使用换元积分法

= \int_{0}^{1} \frac{1}{7 ( v ^{2} - 1 )} d v

提出常数:

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

= \frac{1}{7} \cdot \int_{0}^{1} \frac{1}{v ^{2} - 1} d v

因式分解

v^{2} - 1 : - (- v^{2} + 1)

= \frac{1}{7} \cdot \int_{0}^{1} \frac{1}{- ( - v ^{2} + 1 )} d v

提出常数:

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

= \frac{1}{7} (- \int_{0}^{1} \frac{1}{- v ^{2} + 1} d v)

使用积分常见定则:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{7} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{1})

化简

\frac{1}{7} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{1}) : - \frac{1}{7} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{1}

= - \frac{1}{7} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{1}

带入上下限计算后得:

发散

= 发散

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