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受欢迎的微积分 >

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

解答

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

解答

发散

求解步骤

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

化简

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

= \int_{0}^{1} \frac{x ^{2} + 1}{x ^{2}} d x

化简

\frac{x ^{2} + 1}{x ^{2}} : \frac{x ^{2} + 1}{x ^{2}}

= \int_{0}^{1} \frac{x ^{2} + 1}{x ^{2}} d x

x^{2} = (x),

假定

x \geq 0

= \int_{0}^{1} \frac{x ^{2} + 1}{x} d x

使用换元积分法

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

因式分解

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

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

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

= - \int_{1}^{2} \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_{1}^{2} \frac{1}{- u ^{2} + 1} d u - \int_{1}^{2} 1 d u)

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

发散

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= 发散

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