integral of 4x^2+1/((2x)^2)

解答

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

\frac{4 x ^{3}}{3} + \frac{1}{4 x} - \frac{1}{2 x} + C

求解步骤

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

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

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

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

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

= \frac{4 x ^{3}}{3} + \frac{1}{4 x} - \frac{1}{2 x}

解答补常数

= \frac{4 x ^{3}}{3} + \frac{1}{4 x} - \frac{1}{2 x} + C

l a pl a ce t r an s f or m 2 t^{4}

x \frac{d y}{d x} = y + 2 x e^{- \frac{y}{x}}

\frac{d y}{d x} = x - 2.5 y

ma c l a u r in 1.5 (1 + e^{- 0.2 t})

x \to 0 lim ((1 - 4 x)^{\frac{6}{x}})