integral of (9x^5-3x+1)/(x^3-x^2)

解答

\int \frac{9 x ^{5} - 3 x + 1}{x ^{3} - x ^{2}} d x

27 x + 3 (x - 1)^{3} + \frac{27}{2} (x - 1)^{2} + 7 ln ∣ x - 1 ∣ + 2 ln ∣ x ∣ + \frac{1}{x} - 27 + C

求解步骤

\int \frac{9 x ^{5} - 3 x + 1}{x ^{3} - x ^{2}} d x

乘开

\frac{9 x ^{5} - 3 x + 1}{x ^{3} - x ^{2}} : \frac{9 x ^{5}}{x ^{3} - x ^{2}} - \frac{3 x}{x ^{3} - x ^{2}} + \frac{1}{x ^{3} - x ^{2}}

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{9 x ^{5}}{x ^{3} - x ^{2}} d x - \int \frac{3 x}{x ^{3} - x ^{2}} d x + \int \frac{1}{x ^{3} - x ^{2}} d x

\int \frac{9 x ^{5}}{x ^{3} - x ^{2}} d x = 9 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣)

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

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

= 9 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣) - 3 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) - ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣

化简

9 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣) - 3 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) - ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣ : 27 x + 3 (x - 1)^{3} + \frac{27}{2} (x - 1)^{2} + 7 ln ∣ x - 1 ∣ + 2 ln ∣ x ∣ + \frac{1}{x} - 27

= 27 x + 3 (x - 1)^{3} + \frac{27}{2} (x - 1)^{2} + 7 ln ∣ x - 1 ∣ + 2 ln ∣ x ∣ + \frac{1}{x} - 27

解答补常数

= 27 x + 3 (x - 1)^{3} + \frac{27}{2} (x - 1)^{2} + 7 ln ∣ x - 1 ∣ + 2 ln ∣ x ∣ + \frac{1}{x} - 27 + C

\int 6 x - 18 d x

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