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 1/((x^2+2)^{5/2)}

解答

\int \frac{1}{( x ^{2} + 2 ) ^{\frac{5}{2}}} d x

解答

\frac{x ( x ^{2} + 3 )}{6 ( x ^{2} + 2 ) ^{\frac{3}{2}}} + C

求解步骤

\int \frac{1}{( x ^{2} + 2 ) ^{\frac{5}{2}}} d x

使用三角换元法

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

提出常数:

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

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

使用三角恒等式改写

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

使用换元积分法

= \frac{1}{4} \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

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

\int 1 d v = v

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

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

代回

= \frac{1}{4} sin (arctan (\frac{1}{2 ^{\frac{1}{2}}} x)) - \frac{sin ^{3} ( arctan ( \frac{1}{2 ^{\frac{1}{2}}} x ) )}{3}

化简

\frac{1}{4} sin (arctan (\frac{1}{2 ^{\frac{1}{2}}} x)) - \frac{sin ^{3} ( arctan ( \frac{1}{2 ^{\frac{1}{2}}} x ) )}{3} : \frac{x ( x ^{2} + 3 )}{6 ( x ^{2} + 2 ) ^{\frac{3}{2}}}

= \frac{x ( x ^{2} + 3 )}{6 ( x ^{2} + 2 ) ^{\frac{3}{2}}}

解答补常数

= \frac{x ( x ^{2} + 3 )}{6 ( x ^{2} + 2 ) ^{\frac{3}{2}}} + C

作图

流行的例子

integral of (u-1)/(u^2)

\int \frac{u - 1}{u ^{2}} d u

integral from 0 to infinity of xe^{-x^2}

\int_{0}^{\infty} x e^{- x^{2}} d x

(\partial)/(\partial y)(1/(x-y))

\frac{\partial}{\partial y} (\frac{1}{x - y})

inverselaplace 2

in v erse l a pl a ce 2

tangent f(x)= 9/(x^2+5)

t an g e n t f (x) = \frac{9}{x ^{2} + 5}