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 (16)/(16-x^2)

解答

\int \frac{16}{16 - x ^{2}} d x

解答

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

求解步骤

\int \frac{16}{16 - x ^{2}} d x

提出常数:

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

= 16 \cdot \int \frac{1}{16 - x ^{2}} d x

使用换元积分法

= 16 \cdot \int \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

= 16 \cdot \frac{1}{4} \cdot \int \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}

= 16 \cdot \frac{1}{4} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

u = \frac{x}{4}

代回

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

化简

16 \cdot \frac{1}{4} (\frac{ln \frac{x}{4} + 1}{2} - \frac{ln \frac{x}{4} - 1}{2}) : 4 (\frac{1}{2} ln \frac{x}{4} + 1 - \frac{1}{2} ln \frac{x}{4} - 1)

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

解答补常数

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

作图

流行的例子

derivative (2x+6)/(x+9)

d er i v a t i v e \frac{2 x + 6}{x + 9}

limit as x approaches 0 of 1/x-cot(x)

x \to 0 lim (\frac{1}{x} - cot (x))

integral of x*e^{3x^2}

\int x \cdot e^{3 x^{2}} d x

derivative of tan(x^3+1)

\frac{d}{d x} (tan (x^{3} + 1))

derivative f(x)=(-x^2+3)^2

d er i v a t i v e f (x) = (- x^{2} + 3)^{2}