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受欢迎的微积分 >

integral from 0 to 1 of (24)/(x^2-36)

解答

\int_{0}^{1} \frac{24}{x ^{2} - 36} d x

解答

- 2 (ln (\frac{7}{6}) - ln (\frac{5}{6}))

+1

十进制

- 0.67294 \dots

求解步骤

\int_{0}^{1} \frac{24}{x ^{2} - 36} d x

提出常数:

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

= 24 \cdot \int_{0}^{1} \frac{1}{x ^{2} - 36} d x

使用换元积分法

= 24 \cdot \int_{0}^{\frac{1}{6}} \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

= 24 \cdot \frac{1}{6} \cdot \int_{0}^{\frac{1}{6}} \frac{1}{u ^{2} - 1} d u

因式分解

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

= 24 \cdot \frac{1}{6} \cdot \int_{0}^{\frac{1}{6}} \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

= 24 \cdot \frac{1}{6} (- \int_{0}^{\frac{1}{6}} \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}

= 24 \cdot \frac{1}{6} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{6}})

化简

24 \cdot \frac{1}{6} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{6}}) : - 4 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{\frac{1}{6}}

= - 4 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{\frac{1}{6}}

带入上下限计算后得:

\frac{ln ( \frac{7}{6} ) - ln ( \frac{5}{6} )}{2}

= - 4 \cdot \frac{ln ( \frac{7}{6} ) - ln ( \frac{5}{6} )}{2}

化简

= - 2 (ln (\frac{7}{6}) - ln (\frac{5}{6}))

作图

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