zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

integral from 1 to 2 of 1/(xsqrt(4-x^2))

解答

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

解答

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

+1

十进制

0.65847 \dots

求解步骤

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

使用换元积分法

= \int_{3}^{0} - \frac{1}{- u ^{2} + 4} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{3} - \frac{1}{- u ^{2} + 4} d u

提出常数:

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

= - (- \int_{0}^{3} \frac{1}{- u ^{2} + 4} d u)

使用换元积分法

= - (- \int_{0}^{\frac{3}{2}} \frac{1}{2 ( - v ^{2} + 1 )} d v)

提出常数:

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

= - (- \frac{1}{2} \cdot \int_{0}^{\frac{3}{2}} \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}

= - - \frac{1}{2} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{\frac{3}{2}}

化简

- - \frac{1}{2} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{\frac{3}{2}} : \frac{1}{2} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{\frac{3}{2}}

= \frac{1}{2} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{\frac{3}{2}}

带入上下限计算后得:

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

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

化简

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

作图

流行的例子

integral from 1 to 2 of (2-t)e^{-st}

\int_{1}^{2} (2 - t) e^{- s t} d t

integral from 1 to e of (t^5+1)/t

\int_{1}^{e} \frac{t ^{5} + 1}{t} d t

integral from 0 to 300 of 2x+500

\int_{0}^{300} 2 x + 500 d x

integral from 0 to 4 of |x^2-4|

\int_{0}^{4} x^{2} - 4 d x

integral from 1 to 2 of (v^4+4v^8)/(v^5)

\int_{1}^{2} \frac{v ^{4} + 4 v ^{8}}{v ^{5}} d v