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 0 to 4 of 1/(x^2-16)

解答

\int_{0}^{4} \frac{1}{x ^{2} - 16} d x

解答

发散

求解步骤

\int_{0}^{4} \frac{1}{x ^{2} - 16} d x

使用换元积分法

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

提出常数:

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

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

因式分解

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

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

提出常数:

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

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

使用积分常见定则:

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

= \frac{1}{4} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{1})

化简

\frac{1}{4} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{1}) : - \frac{1}{4} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{1}

= - \frac{1}{4} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{1}

带入上下限计算后得:

发散

= 发散

流行的例子

limit as x approaches 4 of (5-x)/(x-4)

x \to 4 lim (\frac{5 - x}{x - 4})

面积 6x,8x^2,(0,0.75)

a re a 6 x, 8 x^{2}, (0, 0.75)

derivative of ce^{5t}+ce^{5{y}(t,x})

\frac{d}{d x} (c e^{5 t} + c e^{5 y (t, x)})

integral from 4.5 to 5 of x/8

\int_{4.5}^{5} \frac{x}{8} d x

1/4 y^{''}+y^'+y=x^2-2x

\frac{1}{4} y^{^{''}} + y^{^{'}} + y = x^{2} - 2 x