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受欢迎的微积分 >

integral from 1 to 2 of (sqrt(x+4))/x

解答

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

解答

- 4 - \frac{3}{2} + \frac{5 + ln ( \frac{3}{2} + 1 ) + ln ( \frac{5}{2} - 1 ) - ln ( - \frac{5}{2} + \frac{3 ( 2 + 5 )}{2 2} - 1 )}{2}

+1

十进制

1.61652 \dots

求解步骤

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

使用换元积分法

= \int_{5}^{6} \frac{2 u ^{2}}{u ^{2} - 4} d u

提出常数:

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

= 2 \cdot \int_{5}^{6} \frac{u ^{2}}{u ^{2} - 4} d u

因式分解

u^{2} - 4 : - (- u^{2} + 4)

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

提出常数:

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

= 2 (- \int_{5}^{6} \frac{u ^{2}}{- u ^{2} + 4} d u)

使用换元积分法

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

提出常数:

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

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

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= 2 - 2 \frac{ln ( \frac{3}{2} + 1 ) - ln ( \frac{3 ( 2 + 5 )}{2 2} - \frac{5}{2} - 1 ) + ln ( \frac{5}{2} - 1 )}{2} - (\frac{3}{2} - \frac{5}{2})

化简

2 - 2 \frac{ln ( \frac{3}{2} + 1 ) - ln ( \frac{3 ( 2 + 5 )}{2 2} - \frac{5}{2} - 1 ) + ln ( \frac{5}{2} - 1 )}{2} - (\frac{3}{2} - \frac{5}{2}) : - 4 - \frac{3}{2} + \frac{5 + ln ( \frac{3}{2} + 1 ) + ln ( \frac{5}{2} - 1 ) - ln ( - \frac{5}{2} + \frac{3 ( 2 + 5 )}{2 2} - 1 )}{2}

= - 4 - \frac{3}{2} + \frac{5 + ln ( \frac{3}{2} + 1 ) + ln ( \frac{5}{2} - 1 ) - ln ( - \frac{5}{2} + \frac{3 ( 2 + 5 )}{2 2} - 1 )}{2}

作图

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