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受欢迎的微积分 >

integral from e to 3 of 1/(16-x^2)

解答

\int_{e}^{3} \frac{1}{16 - x ^{2}} d x

解答

\frac{ln ( 7 ) - ln ( \frac{e}{4} + 1 ) + ln ( - \frac{e}{4} + 1 )}{8}

+1

十进制

0.03615 \dots

求解步骤

\int_{e}^{3} \frac{1}{16 - x ^{2}} d x

使用换元积分法

= \int_{\frac{e}{4}}^{\frac{3}{4}} \frac{1}{4 ( - 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}{4} \cdot \int_{\frac{e}{4}}^{\frac{3}{4}} \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}{4} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{e}{4}}^{\frac{3}{4}}

化简

\frac{1}{4} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{e}{4}}^{\frac{3}{4}} : \frac{1}{4} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{e}{4}}^{\frac{3}{4}}

= \frac{1}{4} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{e}{4}}^{\frac{3}{4}}

带入上下限计算后得:

\frac{ln ( \frac{7}{4} ) + 2 ln ( 2 ) - ln ( \frac{e}{4} + 1 ) + ln ( - \frac{e}{4} + 1 )}{2}

= \frac{1}{4} \cdot \frac{ln ( \frac{7}{4} ) + 2 ln ( 2 ) - ln ( \frac{e}{4} + 1 ) + ln ( - \frac{e}{4} + 1 )}{2}

化简

= \frac{ln ( 7 ) - ln ( \frac{e}{4} + 1 ) + ln ( - \frac{e}{4} + 1 )}{8}

作图

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