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 of 1/((x-1)sqrt(15+2x-x^2))

解答

\int \frac{1}{( x - 1 ) 15 + 2 x - x ^{2}} d x

解答

- \frac{1}{8} (ln \frac{- x ^{2} + 2 x + 15}{4} + 1 - ln \frac{- x ^{2} + 2 x + 15}{4} - 1) + C

求解步骤

\int \frac{1}{( x - 1 ) 15 + 2 x - x ^{2}} d x

使用换元积分法

= \int \frac{1}{u - u ^{2} + 16} d u

使用换元积分法

= \int - \frac{1}{- v ^{2} + 16} d v

提出常数:

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

= - \int \frac{1}{- v ^{2} + 16} d v

使用换元积分法

= - \int \frac{1}{4 ( - w ^{2} + 1 )} d w

提出常数:

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

= - \frac{1}{4} \cdot \int \frac{1}{- w ^{2} + 1} d w

使用积分常见定则:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

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

代回

= - \frac{1}{4} \frac{ln \frac{- ( x - 1 ) ^{2} + 16}{4} + 1}{2} - \frac{ln \frac{- ( x - 1 ) ^{2} + 16}{4} - 1}{2}

化简

- \frac{1}{4} \frac{ln \frac{- ( x - 1 ) ^{2} + 16}{4} + 1}{2} - \frac{ln \frac{- ( x - 1 ) ^{2} + 16}{4} - 1}{2} : - \frac{1}{8} (ln \frac{- x ^{2} + 2 x + 15}{4} + 1 - ln \frac{- x ^{2} + 2 x + 15}{4} - 1)

= - \frac{1}{8} (ln \frac{- x ^{2} + 2 x + 15}{4} + 1 - ln \frac{- x ^{2} + 2 x + 15}{4} - 1)

解答补常数

= - \frac{1}{8} (ln \frac{- x ^{2} + 2 x + 15}{4} + 1 - ln \frac{- x ^{2} + 2 x + 15}{4} - 1) + C

作图

流行的例子

integral from 1 to 2 of x^4

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

integral of (1-x^2)^{1/2}

\int (1 - x^{2})^{\frac{1}{2}} d x

(\partial)/(\partial y)(4x^3)

\frac{\partial}{\partial y} (4 x^{3})

limit as x approaches 2+of x^3-cx

x \to 2 + lim (x^{3} - c x)

integral of (e^{1/(x^5)})/(x^6)

\int \frac{e ^{\frac{1}{x ^{5}}}}{x ^{6}} d x