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受欢迎的微积分 >

integral from 0 to 1 of 6/(x^2+2x-8)

解答

\int_{0}^{1} \frac{6}{x ^{2} + 2 x - 8} d x

解答

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

+1

十进制

- 0.91629 \dots

求解步骤

\int_{0}^{1} \frac{6}{x ^{2} + 2 x - 8} d x

提出常数:

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

= 6 \cdot \int_{0}^{1} \frac{1}{x ^{2} + 2 x - 8} d x

因式分解

x^{2} + 2 x - 8 : - (- x^{2} - 2 x + 8)

= 6 \cdot \int_{0}^{1} \frac{1}{- ( - x ^{2} - 2 x + 8 )} d x

提出常数:

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

= 6 (- \int_{0}^{1} \frac{1}{- x ^{2} - 2 x + 8} d x)

对

- x^{2} - 2 x + 8

配方:

- (x + 1)^{2} + 9

= 6 (- \int_{0}^{1} \frac{1}{- ( x + 1 ) ^{2} + 9} d x)

使用换元积分法

= 6 (- \int_{1}^{2} \frac{1}{- u ^{2} + 9} d u)

使用换元积分法

= 6 (- \int_{\frac{1}{3}}^{\frac{2}{3}} \frac{1}{3 ( - v ^{2} + 1 )} d v)

提出常数:

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

= 6 (- \frac{1}{3} \cdot \int_{\frac{1}{3}}^{\frac{2}{3}} \frac{1}{- v ^{2} + 1} d v)

使用积分常见定则:

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

= 6 (- \frac{1}{3} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{1}{3}}^{\frac{2}{3}})

化简

6 (- \frac{1}{3} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{1}{3}}^{\frac{2}{3}}) : - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{1}{3}}^{\frac{2}{3}}

= - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{1}{3}}^{\frac{2}{3}}

带入上下限计算后得:

\frac{ln ( \frac{5}{3} ) + ln ( 3 ) - ln ( \frac{4}{3} ) + ln ( \frac{2}{3} )}{2}

= - 2 \cdot \frac{ln ( \frac{5}{3} ) + ln ( 3 ) - ln ( \frac{4}{3} ) + ln ( \frac{2}{3} )}{2}

化简

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

作图

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