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受欢迎的微积分 >

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

解答

\int_{2}^{4} \frac{2}{x ^{2} + x - 2} d x

解答

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

+1

十进制

0.46209 \dots

求解步骤

\int_{2}^{4} \frac{2}{x ^{2} + x - 2} d x

提出常数:

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

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

对

x^{2} + x - 2

配方:

(x + \frac{1}{2})^{2} - \frac{9}{4}

= 2 \cdot \int_{2}^{4} \frac{1}{( x + \frac{1}{2} ) ^{2} - \frac{9}{4}} d x

使用换元积分法

= 2 \cdot \int_{\frac{5}{2}}^{\frac{9}{2}} \frac{4}{4 u ^{2} - 9} d u

提出常数:

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

= 2 \cdot 4 \cdot \int_{\frac{5}{2}}^{\frac{9}{2}} \frac{1}{4 u ^{2} - 9} d u

使用换元积分法

= 2 \cdot 4 \cdot \int_{\frac{5}{3}}^{3} \frac{1}{6 ( v ^{2} - 1 )} d v

提出常数:

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

= 2 \cdot 4 \cdot \frac{1}{6} \cdot \int_{\frac{5}{3}}^{3} \frac{1}{v ^{2} - 1} d v

因式分解

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

= 2 \cdot 4 \cdot \frac{1}{6} \cdot \int_{\frac{5}{3}}^{3} \frac{1}{- ( - v ^{2} + 1 )} d v

提出常数:

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

= 2 \cdot 4 \cdot \frac{1}{6} (- \int_{\frac{5}{3}}^{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}

= 2 \cdot 4 \cdot \frac{1}{6} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{5}{3}}^{3})

化简

2 \cdot 4 \cdot \frac{1}{6} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{5}{3}}^{3}) : - \frac{4}{3} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{\frac{5}{3}}^{3}

= - \frac{4}{3} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{\frac{5}{3}}^{3}

带入上下限计算后得:

\frac{ln ( 2 ) - ln ( \frac{8}{3} ) + ln ( \frac{2}{3} )}{2}

= - \frac{4}{3} \cdot \frac{ln ( 2 ) - ln ( \frac{8}{3} ) + ln ( \frac{2}{3} )}{2}

化简

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

作图

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