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 (u^2)/(sqrt(u^2-8u))

解答

\int \frac{u ^{2}}{u ^{2} - 8 u} d u

解答

\frac{1}{2} (u - 4) u^{2} - 8 u + 24 ln (\frac{1}{4} u + u^{2} - 8 u - 4) + 8 u^{2} - 8 u + C

求解步骤

\int \frac{u ^{2}}{u ^{2} - 8 u} d u

对

u^{2} - 8 u

配方:

(u - 4)^{2} - 16

= \int \frac{u ^{2}}{( u - 4 ) ^{2} - 16} d u

使用换元积分法

= \int \frac{( v + 4 ) ^{2}}{v ^{2} - 16} d v

乘开

\frac{( v + 4 ) ^{2}}{v ^{2} - 16} : \frac{v ^{2}}{v ^{2} - 16} + \frac{8 v}{v ^{2} - 16} + \frac{16}{v ^{2} - 16}

= \int \frac{v ^{2}}{v ^{2} - 16} + \frac{8 v}{v ^{2} - 16} + \frac{16}{v ^{2} - 16} 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

= \int \frac{v ^{2}}{v ^{2} - 16} d v + \int \frac{8 v}{v ^{2} - 16} d v + \int \frac{16}{v ^{2} - 16} d v

\int \frac{v ^{2}}{v ^{2} - 16} d v = 16 (\frac{1}{32} v v^{2} - 16 + \frac{1}{2} ln (\frac{v ^{2} - 16 + v}{4}))

\int \frac{8 v}{v ^{2} - 16} d v = 8 v^{2} - 16

\int \frac{16}{v ^{2} - 16} d v = 16 ln (\frac{1}{4} v^{2} - 16 + v)

= 16 (\frac{1}{32} v v^{2} - 16 + \frac{1}{2} ln (\frac{v ^{2} - 16 + v}{4})) + 8 v^{2} - 16 + 16 ln (\frac{1}{4} v^{2} - 16 + v)

v = u - 4

代回

= 16 \frac{1}{32} (u - 4) (u - 4)^{2} - 16 + \frac{1}{2} ln \frac{( u - 4 ) ^{2} - 16 + u - 4}{4} + 8 (u - 4)^{2} - 16 + 16 ln (\frac{1}{4} (u - 4)^{2} - 16 + u - 4)

化简

16 \frac{1}{32} (u - 4) (u - 4)^{2} - 16 + \frac{1}{2} ln \frac{( u - 4 ) ^{2} - 16 + u - 4}{4} + 8 (u - 4)^{2} - 16 + 16 ln (\frac{1}{4} (u - 4)^{2} - 16 + u - 4) : \frac{1}{2} (u - 4) u^{2} - 8 u + 24 ln (\frac{1}{4} u + u^{2} - 8 u - 4) + 8 u^{2} - 8 u

= \frac{1}{2} (u - 4) u^{2} - 8 u + 24 ln (\frac{1}{4} u + u^{2} - 8 u - 4) + 8 u^{2} - 8 u

解答补常数

= \frac{1}{2} (u - 4) u^{2} - 8 u + 24 ln (\frac{1}{4} u + u^{2} - 8 u - 4) + 8 u^{2} - 8 u + C

作图

流行的例子

derivative of x^{1/3}(x+4)

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

derivative 2/(1+3x+x^2)

d er i v a t i v e \frac{2}{1 + 3 x + x ^{2}}

(24+x^2)y^'-2xy=0

(24 + x^{2}) y^{^{'}} - 2 x y = 0

(dy}{dx}=\frac{y^3)/x ,y(1)=1

\frac{d y}{d x} = \frac{y ^{3}}{x}, y (1) = 1

integral of (x^2-5x+9)/(x^2-5x+6)

\int \frac{x ^{2} - 5 x + 9}{x ^{2} - 5 x + 6} d x