integral of (-5x^2)/(x^4-1)

解答

\int \frac{- 5 x ^{2}}{x ^{4} - 1} d x

- 5 (\frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) + C

求解步骤

\int \frac{- 5 x ^{2}}{x ^{4} - 1} d x

提出常数:

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

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

将

\frac{x ^{2}}{x ^{4} - 1}

用部份分式展开:

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

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

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

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

= - 5 (\frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣)

解答补常数

= - 5 (\frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) + C

\frac{\partial}{\partial t} (cos^{2} (t))

x \to 0 - lim (2 x + c)

\frac{d y}{d x} = \frac{x ^{2} - 1}{y ^{2}}

y^{^{'''}} + 4 y^{^{''}} - 5 y^{^{'}} = 0

\int_{0}^{8} (8 x - x^{2}) d x