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

解答

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

\frac{8}{33} ln ∣ 3 x - 2 ∣ + \frac{1}{11} (\frac{1}{2} ln x^{2} + 2 - \frac{3}{2} arctan (\frac{x}{2})) + C

求解步骤

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

将

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

用部份分式展开:

\frac{8}{11 ( 3 x - 2 )} + \frac{x - 3}{11 ( x ^{2} + 2 )}

= \int \frac{8}{11 ( 3 x - 2 )} + \frac{x - 3}{11 ( x ^{2} + 2 )} 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{8}{11 ( 3 x - 2 )} d x + \int \frac{x - 3}{11 ( x ^{2} + 2 )} d x

\int \frac{8}{11 ( 3 x - 2 )} d x = \frac{8}{33} ln ∣ 3 x - 2 ∣

\int \frac{x - 3}{11 ( x ^{2} + 2 )} d x = \frac{1}{11} (\frac{1}{2} ln x^{2} + 2 - \frac{3}{2} arctan (\frac{x}{2}))

= \frac{8}{33} ln ∣ 3 x - 2 ∣ + \frac{1}{11} (\frac{1}{2} ln x^{2} + 2 - \frac{3}{2} arctan (\frac{x}{2}))

解答补常数

= \frac{8}{33} ln ∣ 3 x - 2 ∣ + \frac{1}{11} (\frac{1}{2} ln x^{2} + 2 - \frac{3}{2} arctan (\frac{x}{2})) + C

a re a y = \frac{x ^{2}}{2} + 2, [- 3, 0]

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

a re a y = - x^{2} - 2 x + 3, y = - x + 1

\int \frac{( x - 1 )}{( x ^{2} - 2 x - 35 ) ^{7}} d x

d er i v a t i v e \frac{9}{t}