integral of ((-3x^3+\sqrt[5]{x}))/(5x)

解答

\int \frac{( - 3 x ^{3} + 5 x )}{5 x} d x

- \frac{1}{5} x^{3} + 5 x + C

求解步骤

\int \frac{- 3 x ^{3} + 5 x}{5 x} d x

提出常数:

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

= \frac{1}{5} \cdot \int \frac{- 3 x ^{3} + 5 x}{x} d x

乘开

\frac{- 3 x ^{3} + 5 x}{x} : - 3 x^{2} + \frac{1}{x ^{\frac{4}{5}}}

= \frac{1}{5} \cdot \int - 3 x^{2} + \frac{1}{x ^{\frac{4}{5}}} 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

= \frac{1}{5} (- \int 3 x^{2} d x + \int \frac{1}{x ^{\frac{4}{5}}} d x)

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

\int \frac{1}{x ^{\frac{4}{5}}} d x = 5 x^{\frac{1}{5}}

= \frac{1}{5} (- x^{3} + 5 x^{\frac{1}{5}})

化简

\frac{1}{5} (- x^{3} + 5 x^{\frac{1}{5}}) : - \frac{1}{5} x^{3} + 5 x

= - \frac{1}{5} x^{3} + 5 x

解答补常数

= - \frac{1}{5} x^{3} + 5 x + C

\frac{d}{d x} (e^{- ∣ x ∣})

x \to - 2 lim (\frac{x + 2}{2 x + 4})

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

a re a y = 4 x^{2} - 1, y = 1 - 4 x^{2}

y^{'} = y (2 - y)