integral of (2u^2-2u^{-1})

解答

\int (2 u^{2} - 2 u^{- 1}) d u

\frac{2 u ^{3}}{3} - 2 ln ∣ u ∣ + C

求解步骤

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

2 u^{- 1} = \frac{2}{u}

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

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

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

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= \frac{2 u ^{3}}{3} - 2 ln ∣ u ∣

解答补常数

= \frac{2 u ^{3}}{3} - 2 ln ∣ u ∣ + C

3 y^{^{''}} + 2 y^{^{'}} + y = 0

\frac{\partial}{\partial y} (\frac{y}{( x ^{2} + y ^{2} )})

d er i v a t i v e f (x) = 4 x^{6} - 7 x^{5}

d er i v a t i v e 5 x sin (π x)

t \to - 2 lim (\frac{t ^{3} + 8}{t + 2})