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^3-4x)

解答

\int \frac{1}{x ^{3} - 4 x} d x

解答

- \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣ + C

求解步骤

\int \frac{1}{x ^{3} - 4 x} d x

因式分解

x^{3} - 4 x : - (- x^{3} + 4 x)

= \int \frac{1}{- ( - x ^{3} + 4 x )} d x

提出常数:

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

= - \int \frac{1}{- x ^{3} + 4 x} d x

将

\frac{1}{- x ^{3} + 4 x}

用部份分式展开:

\frac{1}{4 x} - \frac{1}{8 ( x + 2 )} - \frac{1}{8 ( x - 2 )}

= - \int \frac{1}{4 x} - \frac{1}{8 ( x + 2 )} - \frac{1}{8 ( x - 2 )} d x

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}{4 x} d x - \int \frac{1}{8 ( x + 2 )} d x - \int \frac{1}{8 ( x - 2 )} d x)

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

\int \frac{1}{8 ( x + 2 )} d x = \frac{1}{8} ln ∣ x + 2 ∣

\int \frac{1}{8 ( x - 2 )} d x = \frac{1}{8} ln ∣ x - 2 ∣

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

化简

= - \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣

解答补常数

= - \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣ + C

作图

流行的例子

integral of (3x^2+1)/((x^2-1)^3)

\int \frac{3 x ^{2} + 1}{( x ^{2} - 1 ) ^{3}} d x

derivative 8sin(8x-1)

d er i v a t i v e 8 sin (8 x - 1)

derivative of e^x(2x+3^3)

\frac{d}{d x} (e^{x} (2 x + 3)^{3})

(d^2)/(dx^2)(e^{3/x})

\frac{d ^{2}}{d x ^{2}} (e^{\frac{3}{x}})

integral of (9x)/(x^2+1)

\int \frac{9 x}{x ^{2} + 1} d x