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

解答

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

3 x + \frac{1}{2} (x - 1)^{2} + 4 ln ∣ x - 1 ∣ - \frac{3}{x - 1} - 3 + C

求解步骤

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

乘开

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

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

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

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

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

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

化简

\frac{( x - 1 ) ^{2}}{2} + 3 (x - 1) + 3 ln ∣ x - 1 ∣ - \frac{1}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1} - \frac{1}{x - 1} : 3 x + \frac{1}{2} (x - 1)^{2} + 4 ln ∣ x - 1 ∣ - \frac{3}{x - 1} - 3

= 3 x + \frac{1}{2} (x - 1)^{2} + 4 ln ∣ x - 1 ∣ - \frac{3}{x - 1} - 3

解答补常数

= 3 x + \frac{1}{2} (x - 1)^{2} + 4 ln ∣ x - 1 ∣ - \frac{3}{x - 1} - 3 + C

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