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 4/(x^4sqrt(x^2-8))

解答

\int \frac{4}{x ^{4} x ^{2} - 8} d x

解答

\frac{( x ^{2} + 4 ) x ^{2} - 8}{24 x ^{3}} + C

求解步骤

\int \frac{4}{x ^{4} x ^{2} - 8} d x

提出常数:

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

= 4 \cdot \int \frac{1}{x ^{4} x ^{2} - 8} d x

使用三角换元法

= 4 \cdot \int \frac{1}{64 sec ^{3} ( u )} d u

提出常数:

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

= 4 \cdot \frac{1}{64} \cdot \int \frac{1}{sec ^{3} ( u )} d u

使用三角恒等式改写

= 4 \cdot \frac{1}{64} \cdot \int (1 - sin^{2} (u)) cos (u) d u

使用换元积分法

= 4 \cdot \frac{1}{64} \cdot \int 1 - v^{2} 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

= 4 \cdot \frac{1}{64} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

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

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

代回

= 4 \cdot \frac{1}{64} sin (arcsec (\frac{1}{2 2} x)) - \frac{sin ^{3} ( arcsec ( \frac{1}{2 2} x ) )}{3}

化简

4 \cdot \frac{1}{64} sin (arcsec (\frac{1}{2 2} x)) - \frac{sin ^{3} ( arcsec ( \frac{1}{2 2} x ) )}{3} : \frac{( x ^{2} + 4 ) x ^{2} - 8}{24 x ^{3}}

= \frac{( x ^{2} + 4 ) x ^{2} - 8}{24 x ^{3}}

解答补常数

= \frac{( x ^{2} + 4 ) x ^{2} - 8}{24 x ^{3}} + C

作图

流行的例子

integral of (-3xy^2-2ycos(x))

integral of 3xarcsin(x)

integral of 1/(x(x+1)(x+2)(x+3))

integral of (e^{3x}+3e^x+3)/(e^x)

integral of ((y-3))/(4y)