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^2-1)^3)

解答

\int (x^{2} - 1)^{3} d x

解答

32 (\frac{3}{256} ln tan (\frac{1}{2} arcsec (x)) + 1 - \frac{3}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 )} + \frac{1}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{1}{2} arcsec (x)) - 1 - \frac{3}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 )} - \frac{1}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 ) ^{4}}) + C

求解步骤

\int (x^{2} - 1)^{3} d x

(x^{2} - 1)^{3} = (x^{2} - 1)^{\frac{3}{2}},

假定

(x^{2} - 1) \geq 0

= \int (x^{2} - 1)^{\frac{3}{2}} d x

使用三角换元法

= \int tan^{4} (u) sec (u) d u

使用三角恒等式改写

= \int \frac{sin ^{4} ( u )}{cos ^{5} ( u )} d u

使用换元积分法

= \int \frac{32 v ^{4}}{( 1 - v ^{2} ) ^{5}} d v

提出常数:

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

= 32 \cdot \int \frac{v ^{4}}{( 1 - v ^{2} ) ^{5}} d v

将

\frac{v ^{4}}{( 1 - v ^{2} ) ^{5}}

用部份分式展开:

\frac{3}{256 ( v + 1 )} + \frac{3}{256 ( v + 1 ) ^{2}} - \frac{1}{128 ( v + 1 ) ^{3}} - \frac{3}{64 ( v + 1 ) ^{4}} + \frac{1}{32 ( v + 1 ) ^{5}} - \frac{3}{256 ( v - 1 )} + \frac{3}{256 ( v - 1 ) ^{2}} + \frac{1}{128 ( v - 1 ) ^{3}} - \frac{3}{64 ( v - 1 ) ^{4}} - \frac{1}{32 ( v - 1 ) ^{5}}

= 32 \cdot \int \frac{3}{256 ( v + 1 )} + \frac{3}{256 ( v + 1 ) ^{2}} - \frac{1}{128 ( v + 1 ) ^{3}} - \frac{3}{64 ( v + 1 ) ^{4}} + \frac{1}{32 ( v + 1 ) ^{5}} - \frac{3}{256 ( v - 1 )} + \frac{3}{256 ( v - 1 ) ^{2}} + \frac{1}{128 ( v - 1 ) ^{3}} - \frac{3}{64 ( v - 1 ) ^{4}} - \frac{1}{32 ( v - 1 ) ^{5}} 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

= 32 (\int \frac{3}{256 ( v + 1 )} d v + \int \frac{3}{256 ( v + 1 ) ^{2}} d v - \int \frac{1}{128 ( v + 1 ) ^{3}} d v - \int \frac{3}{64 ( v + 1 ) ^{4}} d v + \int \frac{1}{32 ( v + 1 ) ^{5}} d v - \int \frac{3}{256 ( v - 1 )} d v + \int \frac{3}{256 ( v - 1 ) ^{2}} d v + \int \frac{1}{128 ( v - 1 ) ^{3}} d v - \int \frac{3}{64 ( v - 1 ) ^{4}} d v - \int \frac{1}{32 ( v - 1 ) ^{5}} d v)

\int \frac{3}{256 ( v + 1 )} d v = \frac{3}{256} ln ∣ v + 1 ∣

\int \frac{3}{256 ( v + 1 ) ^{2}} d v = - \frac{3}{256 ( v + 1 )}

\int \frac{1}{128 ( v + 1 ) ^{3}} d v = - \frac{1}{256 ( v + 1 ) ^{2}}

\int \frac{3}{64 ( v + 1 ) ^{4}} d v = - \frac{1}{64 ( v + 1 ) ^{3}}

\int \frac{1}{32 ( v + 1 ) ^{5}} d v = - \frac{1}{128 ( v + 1 ) ^{4}}

\int \frac{3}{256 ( v - 1 )} d v = \frac{3}{256} ln ∣ v - 1 ∣

\int \frac{3}{256 ( v - 1 ) ^{2}} d v = - \frac{3}{256 ( v - 1 )}

\int \frac{1}{128 ( v - 1 ) ^{3}} d v = - \frac{1}{256 ( v - 1 ) ^{2}}

\int \frac{3}{64 ( v - 1 ) ^{4}} d v = - \frac{1}{64 ( v - 1 ) ^{3}}

\int \frac{1}{32 ( v - 1 ) ^{5}} d v = - \frac{1}{128 ( v - 1 ) ^{4}}

= 32 (\frac{3}{256} ln ∣ v + 1 ∣ - \frac{3}{256 ( v + 1 )} - (- \frac{1}{256 ( v + 1 ) ^{2}}) - (- \frac{1}{64 ( v + 1 ) ^{3}}) - \frac{1}{128 ( v + 1 ) ^{4}} - \frac{3}{256} ln ∣ v - 1 ∣ - \frac{3}{256 ( v - 1 )} - \frac{1}{256 ( v - 1 ) ^{2}} - (- \frac{1}{64 ( v - 1 ) ^{3}}) - (- \frac{1}{128 ( v - 1 ) ^{4}}))

代回

= 32 \frac{3}{256} ln tan (\frac{arcsec ( x )}{2}) + 1 - \frac{3}{256 ( tan ( \frac{arcsec ( x )}{2} ) + 1 )} - - \frac{1}{256 ( tan ( \frac{arcsec ( x )}{2} ) + 1 ) ^{2}} - - \frac{1}{64 ( tan ( \frac{arcsec ( x )}{2} ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{arcsec ( x )}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{arcsec ( x )}{2}) - 1 - \frac{3}{256 ( tan ( \frac{arcsec ( x )}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{arcsec ( x )}{2} ) - 1 ) ^{2}} - - \frac{1}{64 ( tan ( \frac{arcsec ( x )}{2} ) - 1 ) ^{3}} - - \frac{1}{128 ( tan ( \frac{arcsec ( x )}{2} ) - 1 ) ^{4}}

化简

32 \frac{3}{256} ln tan (\frac{arcsec ( x )}{2}) + 1 - \frac{3}{256 ( tan ( \frac{arcsec ( x )}{2} ) + 1 )} - - \frac{1}{256 ( tan ( \frac{arcsec ( x )}{2} ) + 1 ) ^{2}} - - \frac{1}{64 ( tan ( \frac{arcsec ( x )}{2} ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{arcsec ( x )}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{arcsec ( x )}{2}) - 1 - \frac{3}{256 ( tan ( \frac{arcsec ( x )}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{arcsec ( x )}{2} ) - 1 ) ^{2}} - - \frac{1}{64 ( tan ( \frac{arcsec ( x )}{2} ) - 1 ) ^{3}} - - \frac{1}{128 ( tan ( \frac{arcsec ( x )}{2} ) - 1 ) ^{4}} : 32 (\frac{3}{256} ln tan (\frac{1}{2} arcsec (x)) + 1 - \frac{3}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 )} + \frac{1}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{1}{2} arcsec (x)) - 1 - \frac{3}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 )} - \frac{1}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 ) ^{4}})

= 32 (\frac{3}{256} ln tan (\frac{1}{2} arcsec (x)) + 1 - \frac{3}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 )} + \frac{1}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{1}{2} arcsec (x)) - 1 - \frac{3}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 )} - \frac{1}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 ) ^{4}})

解答补常数

= 32 (\frac{3}{256} ln tan (\frac{1}{2} arcsec (x)) + 1 - \frac{3}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 )} + \frac{1}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{1}{2} arcsec ( x ) ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{1}{2} arcsec (x)) - 1 - \frac{3}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 )} - \frac{1}{256 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{1}{2} arcsec ( x ) ) - 1 ) ^{4}}) + C

作图

流行的例子

integral of ((2x-3))/((x-1)^2(x-2)^2)

\int \frac{( 2 x - 3 )}{( x - 1 ) ^{2} ( x - 2 ) ^{2}} d x

integral of e^{-5/6 x}

\int e^{- \frac{5}{6} x} d x

integral of vsin(v)

\int v sin (v) d v

integral of 16e^{4x}

\int 16 e^{4 x} d x

integral of (8x^3-12x^2+7x+2)/(2x-2)

\int \frac{8 x ^{3} - 12 x ^{2} + 7 x + 2}{2 x - 2} d x