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

解答

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

\frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} + 8 x^{2} + 68 x - \frac{1}{4} ln ∣ x ∣ + \frac{1}{x} + \frac{1089}{4} ln ∣ x - 4 ∣ + C

求解步骤

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

长除

\frac{x ^{6} + 4 x ^{3} + 4}{x ^{3} - 4 x ^{2}} : x^{3} + 4 x^{2} + 16 x + 68 + \frac{272 x ^{2} + 4}{x ^{3} - 4 x ^{2}}

= \int x^{3} + 4 x^{2} + 16 x + 68 + \frac{272 x ^{2} + 4}{x ^{3} - 4 x ^{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 x^{3} d x + \int 4 x^{2} d x + \int 16 x d x + \int 68 d x + \int \frac{272 x ^{2} + 4}{x ^{3} - 4 x ^{2}} d x

\int x^{3} d x = \frac{x ^{4}}{4}

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 16 x d x = 8 x^{2}

\int 68 d x = 68 x

\int \frac{272 x ^{2} + 4}{x ^{3} - 4 x ^{2}} d x = - \frac{1}{4} ln ∣ x ∣ + \frac{1}{x} + \frac{1089}{4} ln ∣ x - 4 ∣

= \frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} + 8 x^{2} + 68 x - \frac{1}{4} ln ∣ x ∣ + \frac{1}{x} + \frac{1089}{4} ln ∣ x - 4 ∣

解答补常数

= \frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} + 8 x^{2} + 68 x - \frac{1}{4} ln ∣ x ∣ + \frac{1}{x} + \frac{1089}{4} ln ∣ x - 4 ∣ + C

\frac{d}{d x} (cos (x) + x^{5})

\frac{d y}{d x} + 5 x y = 4 x

\int \frac{w}{1 - w} d w

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

\int u^{- 1} d u