integral of 1/(y-y^3)

解答

\int \frac{1}{y - y ^{3}} d y

ln ∣ y ∣ - \frac{1}{2} ln ∣ y + 1 ∣ - \frac{1}{2} ln ∣ y - 1 ∣ + C

求解步骤

\int \frac{1}{y - y ^{3}} d y

将

\frac{1}{y - y ^{3}}

用部份分式展开:

\frac{1}{y} - \frac{1}{2 ( y + 1 )} - \frac{1}{2 ( y - 1 )}

= \int \frac{1}{y} - \frac{1}{2 ( y + 1 )} - \frac{1}{2 ( y - 1 )} d y

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 \frac{1}{y} d y - \int \frac{1}{2 ( y + 1 )} d y - \int \frac{1}{2 ( y - 1 )} d y

\int \frac{1}{y} d y = ln ∣ y ∣

\int \frac{1}{2 ( y + 1 )} d y = \frac{1}{2} ln ∣ y + 1 ∣

\int \frac{1}{2 ( y - 1 )} d y = \frac{1}{2} ln ∣ y - 1 ∣

= ln ∣ y ∣ - \frac{1}{2} ln ∣ y + 1 ∣ - \frac{1}{2} ln ∣ y - 1 ∣

解答补常数

= ln ∣ y ∣ - \frac{1}{2} ln ∣ y + 1 ∣ - \frac{1}{2} ln ∣ y - 1 ∣ + C

\int_{0}^{4} 2 π (x) (4 x - x^{2}) d x

\frac{d}{d x} (\frac{1}{x ^{- 5}})

2 \times 2 + 6 \times 1/22 \times 2

\frac{\partial}{\partial x} (y^{- 2} (4 y + 9 x^{2}))

d er i v a t i v e f (x) = (x^{3} + x)^{6}