integral from 2 to 3 of (sqrt(x^2+1))/x

解答

\int_{2}^{3} \frac{x ^{2} + 1}{x} d x

- \frac{ln ( 10 + 1 ) - ln ( 5 2 + 10 - 5 - 1 ) + ln ( 5 - 1 )}{2} + 10 - 5

十进制

1.07997 \dots

求解步骤

\int_{2}^{3} \frac{x ^{2} + 1}{x} d x

使用换元积分法

= \int_{5}^{10} \frac{u ^{2}}{u ^{2} - 1} d u

因式分解

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

= \int_{5}^{10} \frac{u ^{2}}{- ( - u ^{2} + 1 )} d u

提出常数:

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

= - \int_{5}^{10} \frac{u ^{2}}{- u ^{2} + 1} d u

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

= - \int_{5}^{10} \frac{1}{- u ^{2} + 1} - 1 d u

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

= - (\int_{5}^{10} \frac{1}{- u ^{2} + 1} d u - \int_{5}^{10} 1 d u)

\int_{5}^{10} \frac{1}{- u ^{2} + 1} d u = \frac{ln ( 10 + 1 ) - ln ( 5 2 + 10 - 5 - 1 ) + ln ( 5 - 1 )}{2}

\int_{5}^{10} 1 d u = 10 - 5

= - (\frac{ln ( 10 + 1 ) - ln ( 5 2 + 10 - 5 - 1 ) + ln ( 5 - 1 )}{2} - (10 - 5))

化简

= - \frac{ln ( 10 + 1 ) - ln ( 5 2 + 10 - 5 - 1 ) + ln ( 5 - 1 )}{2} + 10 - 5

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