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+4)^2(x-1))

解答

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

解答

- \frac{1}{25} ln ∣ x + 4 ∣ + \frac{1}{5 ( x + 4 )} + \frac{1}{25} ln ∣ x - 1 ∣ + C

求解步骤

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

使用换元积分法

= \int \frac{1}{u ^{2} ( u - 5 )} d u

将

\frac{1}{u ^{2} ( u - 5 )}

用部份分式展开:

- \frac{1}{25 u} - \frac{1}{5 u ^{2}} + \frac{1}{25 ( u - 5 )}

= \int - \frac{1}{25 u} - \frac{1}{5 u ^{2}} + \frac{1}{25 ( u - 5 )} 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 \frac{1}{25 u} d u - \int \frac{1}{5 u ^{2}} d u + \int \frac{1}{25 ( u - 5 )} d u

\int \frac{1}{25 u} d u = \frac{1}{25} ln ∣ u ∣

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

\int \frac{1}{25 ( u - 5 )} d u = \frac{1}{25} ln ∣ u - 5 ∣

= - \frac{1}{25} ln ∣ u ∣ - (- \frac{1}{5 u}) + \frac{1}{25} ln ∣ u - 5 ∣

u = x + 4

代回

= - \frac{1}{25} ln ∣ x + 4 ∣ - (- \frac{1}{5 ( x + 4 )}) + \frac{1}{25} ln ∣ x + 4 - 5 ∣

化简

- \frac{1}{25} ln ∣ x + 4 ∣ - (- \frac{1}{5 ( x + 4 )}) + \frac{1}{25} ln ∣ x + 4 - 5 ∣ : - \frac{1}{25} ln ∣ x + 4 ∣ + \frac{1}{5 ( x + 4 )} + \frac{1}{25} ln ∣ x - 1 ∣

= - \frac{1}{25} ln ∣ x + 4 ∣ + \frac{1}{5 ( x + 4 )} + \frac{1}{25} ln ∣ x - 1 ∣

解答补常数

= - \frac{1}{25} ln ∣ x + 4 ∣ + \frac{1}{5 ( x + 4 )} + \frac{1}{25} ln ∣ x - 1 ∣ + C

作图

流行的例子

limit as x approaches 0 of (e^x)/(x^3)

x \to 0 lim (\frac{e ^{x}}{x ^{3}})

integral of (3x-1)^5

\int (3 x - 1)^{5} d x

limit as x approaches 0+of (x^3)/(ln(x))

x \to 0 + lim (\frac{x ^{3}}{ln ( x )})

laplacetransform t*e^{-3t}

l a pl a ce t r an s f or m t \cdot e^{- 3 t}

(\partial)/(\partial x)(-2xsin(x^2+y^2))

\frac{\partial}{\partial x} (- 2 x sin (x^{2} + y^{2}))