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受欢迎的微积分 >

integral from 0 to-1 of 1/(x^2-4)

解答

\int_{0}^{- 1} \frac{1}{x ^{2} - 4} d x

解答

\frac{ln ( 3 )}{4}

+1

十进制

0.27465 \dots

求解步骤

\int_{0}^{- 1} \frac{1}{x ^{2} - 4} d x

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

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

使用换元积分法

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

因式分解

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

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

化简

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

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

带入上下限计算后得:

- \frac{- ln ( 2 ) - ln ( \frac{3}{2} )}{2}

= \frac{1}{2} (- \frac{- ln ( 2 ) - ln ( \frac{3}{2} )}{2})

化简

= \frac{ln ( 3 )}{4}

作图

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