integral of ((x^2-x+36))/(x^3+6x)

解答

\int \frac{( x ^{2} - x + 36 )}{x ^{3} + 6 x} d x

6 ln ∣ x ∣ - \frac{5}{2} ln x^{2} + 6 - \frac{1}{6} arctan (\frac{x}{6}) + C

求解步骤

\int \frac{x ^{2} - x + 36}{x ^{3} + 6 x} d x

将

\frac{x ^{2} - x + 36}{x ^{3} + 6 x}

用部份分式展开:

\frac{6}{x} + \frac{- 5 x - 1}{x ^{2} + 6}

= \int \frac{6}{x} + \frac{- 5 x - 1}{x ^{2} + 6} 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{6}{x} d x + \int \frac{- 5 x - 1}{x ^{2} + 6} d x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

\int \frac{- 5 x - 1}{x ^{2} + 6} d x = - \frac{5}{2} ln x^{2} + 6 - \frac{1}{6} arctan (\frac{x}{6})

= 6 ln ∣ x ∣ - \frac{5}{2} ln x^{2} + 6 - \frac{1}{6} arctan (\frac{x}{6})

解答补常数

= 6 ln ∣ x ∣ - \frac{5}{2} ln x^{2} + 6 - \frac{1}{6} arctan (\frac{x}{6}) + C

t an g e n t y = 2 sin (x), (\frac{π}{6}, 1)

d er i v a t i v e 1 - (1 - 0.01 x)^{3}

n = 1 \sum \infty n^{\frac{3}{4}}

\int 90 x^{2} sin (2 + 6 x^{3}) d x

1 - y^{2} d x - 1 - x^{2} d y = 0