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受欢迎的微积分 >

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

解答

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

解答

发散

求解步骤

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

因式分解

a^{2} - x^{2} : - (- a^{2} + x^{2})

= \int_{0}^{a} \frac{1}{- ( - a ^{2} + x ^{2} )} d x

提出常数:

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

= - \int_{0}^{a} \frac{1}{- a ^{2} + x ^{2}} d x

使用换元积分法

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

因式分解

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

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

化简

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

= \frac{1}{a} [\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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