integral of (x+3)/(x^2+6x^4)

解答

\int \frac{x + 3}{x ^{2} + 6 x ^{4}} d x

ln ∣ x ∣ - \frac{3}{x} - \frac{1}{2} ln 6 x^{2} + 1 - 36 arctan (6 x) + C

求解步骤

\int \frac{x + 3}{x ^{2} + 6 x ^{4}} d x

将

\frac{x + 3}{x ^{2} + 6 x ^{4}}

用部份分式展开:

\frac{1}{x} + \frac{3}{x ^{2}} + \frac{- 6 x - 18}{6 x ^{2} + 1}

= \int \frac{1}{x} + \frac{3}{x ^{2}} + \frac{- 6 x - 18}{6 x ^{2} + 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

= \int \frac{1}{x} d x + \int \frac{3}{x ^{2}} d x + \int \frac{- 6 x - 18}{6 x ^{2} + 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int \frac{- 6 x - 18}{6 x ^{2} + 1} d x = - \frac{1}{2} ln 6 x^{2} + 1 - 36 arctan (6 x)

= ln ∣ x ∣ - \frac{3}{x} - \frac{1}{2} ln 6 x^{2} + 1 - 36 arctan (6 x)

解答补常数

= ln ∣ x ∣ - \frac{3}{x} - \frac{1}{2} ln 6 x^{2} + 1 - 36 arctan (6 x) + C

\int_{- \infty}^{\infty} 57 cos (π t) d t

y^{^{'}} = 2 x y^{2} + 3 x^{2} y^{2}, y (1) = - 1

\int 2 a x d x

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

\frac{\partial}{\partial x} (\frac{( 3 x - y )}{( x + 2 y )})