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受欢迎的微积分 >

integral from 0 to 6 of 1/(x^2-36)

解答

\int_{0}^{6} \frac{1}{x ^{2} - 36} d x

解答

发散

求解步骤

\int_{0}^{6} \frac{1}{x ^{2} - 36} d x

使用换元积分法

= \int_{0}^{1} \frac{1}{6 ( u ^{2} - 1 )} d u

提出常数:

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

= \frac{1}{6} \cdot \int_{0}^{1} \frac{1}{u ^{2} - 1} d u

因式分解

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

= \frac{1}{6} \cdot \int_{0}^{1} \frac{1}{- ( - u ^{2} + 1 )} d u

提出常数:

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

= \frac{1}{6} (- \int_{0}^{1} \frac{1}{- u ^{2} + 1} d u)

使用积分常见定则:

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

= \frac{1}{6} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{1})

化简

\frac{1}{6} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{1}) : - \frac{1}{6} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{1}

= - \frac{1}{6} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{1}

带入上下限计算后得:

发散

= 发散

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