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

解答

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

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

求解步骤

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

使用换元积分法

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

提出常数:

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

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

使用分布积分法

= - [\frac{- 1 + u - u \cdot \int - \frac{1}{2 u 1 + u} d u}{u}]_{x}^{\frac{1}{4}}

\int - \frac{1}{2 u 1 + u} d u = \frac{1}{2} ln 1 + u + 1 - \frac{1}{2} ln 1 + u - 1

= - [\frac{- 1 + u - u ( \frac{1}{2} ln 1 + u + 1 - \frac{1}{2} ln 1 + u - 1 )}{u}]_{x}^{\frac{1}{4}}

带入上下限计算后得:

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

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

化简

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

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