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 (sqrt(x^2+16))/(2x)

解答

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

解答

- ln \frac{1}{4} x^{2} + 16 + 1 + ln \frac{1}{4} x^{2} + 16 - 1 + \frac{1}{2} x^{2} + 16 + C

求解步骤

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

提出常数:

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

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

使用换元积分法

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

因式分解

u^{2} - 16 : - (- u^{2} + 16)

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

提出常数:

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

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

使用换元积分法

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

提出常数:

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

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

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

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

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

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

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

\int 1 d v = v

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

代回

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

化简

\frac{1}{2} - 4 \frac{ln \frac{x ^{2} + 16}{4} + 1}{2} - \frac{ln \frac{x ^{2} + 16}{4} - 1}{2} - \frac{x ^{2} + 16}{4} : - ln \frac{1}{4} x^{2} + 16 + 1 + ln \frac{1}{4} x^{2} + 16 - 1 + \frac{1}{2} x^{2} + 16

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

解答补常数

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

作图

流行的例子

derivative of e^{3x+9}

\frac{d}{d x} (e^{3 x + 9})

derivative ln(x+3)

d er i v a t i v e ln (x + 3)

derivative 10+(50t)/(2t^2+9)

d er i v a t i v e 10 + \frac{50 t}{2 t ^{2} + 9}

limit as h approaches 0 of (x+h)^3-x^3

h \to 0 lim ((x + h)^{3} - x^{3})

(\partial)/(\partial t)(2e^{sin(x+2ct)})

\frac{\partial}{\partial t} (2 e^{s i n (x + 2 c t)})