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 x^2sqrt(2x-1)

解答

\int x^{2} 2 x - 1 d x

解答

\frac{1}{105} (15 x^{2} + 6 x + 2) (2 x - 1)^{\frac{3}{2}} + C

求解步骤

\int x^{2} 2 x - 1 d x

使用换元积分法

= \int \frac{u ( u + 1 ) ^{2}}{8} d u

提出常数:

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

= \frac{1}{8} \cdot \int u (u + 1)^{2} d u

乘开

u (u + 1)^{2} : u^{\frac{5}{2}} + 2 u^{\frac{3}{2}} + u

= \frac{1}{8} \cdot \int u^{\frac{5}{2}} + 2 u^{\frac{3}{2}} + u d u

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}{8} (\int u^{\frac{5}{2}} d u + \int 2 u^{\frac{3}{2}} d u + \int u d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 2 u^{\frac{3}{2}} d u = \frac{4}{5} u^{\frac{5}{2}}

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{8} (\frac{2}{7} u^{\frac{7}{2}} + \frac{4}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

u = 2 x - 1

代回

= \frac{1}{8} (\frac{2}{7} (2 x - 1)^{\frac{7}{2}} + \frac{4}{5} (2 x - 1)^{\frac{5}{2}} + \frac{2}{3} (2 x - 1)^{\frac{3}{2}})

化简

\frac{1}{8} (\frac{2}{7} (2 x - 1)^{\frac{7}{2}} + \frac{4}{5} (2 x - 1)^{\frac{5}{2}} + \frac{2}{3} (2 x - 1)^{\frac{3}{2}}) : \frac{1}{105} (15 x^{2} + 6 x + 2) (2 x - 1)^{\frac{3}{2}}

= \frac{1}{105} (15 x^{2} + 6 x + 2) (2 x - 1)^{\frac{3}{2}}

解答补常数

= \frac{1}{105} (15 x^{2} + 6 x + 2) (2 x - 1)^{\frac{3}{2}} + C

作图

流行的例子

(dx)/(dt)+t^7x^9+x/t =0

\frac{d x}{d t} + t^{7} x^{9} + \frac{x}{t} = 0

(d^2)/(dx^2)(cos(sin(3θ)))

\frac{d ^{2}}{d x ^{2}} (cos (sin (3 θ)))

tangent f(x)=-(24)/((4x+1)),\at x=-1

t an g e n t f (x) = - \frac{24}{( 4 x + 1 )}, at x = - 1

derivative (6e^w)/(7w)

d er i v a t i v e \frac{6 e ^{w}}{7 w}

derivative of 14^{2x}

\frac{d}{d x} (1 4^{2 x})