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受欢迎的微积分 >

integral of 1/(x^2sqrt(2x^2+32))

解答

\int \frac{1}{x ^{2} 2 x ^{2} + 32} d x

解答

- \frac{16 + x ^{2}}{16 2 x} + C

求解步骤

\int \frac{1}{x ^{2} 2 x ^{2} + 32} d x

\frac{1}{x ^{2} 2 x ^{2} + 32} = \frac{1}{2} \frac{1}{x ^{2} x ^{2} + 16}

= \int \frac{1}{2} \cdot \frac{1}{x ^{2} x ^{2} + 16} d x

提出常数:

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

= \frac{1}{2} \cdot \int \frac{1}{x ^{2} x ^{2} + 16} d x

使用三角换元法

= \frac{1}{2} \cdot \int \frac{sec ( u )}{16 tan ^{2} ( u )} d u

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{16} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

使用三角恒等式改写

= \frac{1}{2} \cdot \frac{1}{16} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

使用换元积分法

= \frac{1}{2} \cdot \frac{1}{16} \cdot \int \frac{1}{v ^{2}} d v

使用幂法则

= \frac{1}{2} \cdot \frac{1}{16} (- \frac{1}{v})

代回

= \frac{1}{2} \cdot \frac{1}{16} (- \frac{1}{sin ( arctan ( \frac{1}{4} x ) )})

化简

\frac{1}{2} \cdot \frac{1}{16} (- \frac{1}{sin ( arctan ( \frac{1}{4} x ) )}) : - \frac{16 + x ^{2}}{16 2 x}

= - \frac{16 + x ^{2}}{16 2 x}

解答补常数

= - \frac{16 + x ^{2}}{16 2 x} + C

作图

流行的例子

64y^{''}+144y^'+81y=0

64 y^{^{''}} + 144 y^{^{'}} + 81 y = 0

inverselaplace (10)/(s^2+s+10)*1/s

in v erse l a pl a ce \frac{10}{s ^{2} + s + 10} \cdot \frac{1}{s}

(\partial)/(\partial x)((e^y)/(x+y^5))

\frac{\partial}{\partial x} (\frac{e ^{y}}{x + y ^{5}})

(\partial)/(\partial x)(cos(2cx+act))

\frac{\partial}{\partial x} (cos (2 c x + a c t))

derivative of 1000(1-x/(60)^2)

\frac{d}{d x} (1000 (1 - \frac{x}{60})^{2})