integral of (x^2+1)/(x^3+10x^2+21x)

解答

\int \frac{x ^{2} + 1}{x ^{3} + 10 x ^{2} + 21 x} d x

- \frac{5}{6} ln ∣ x + 3 ∣ + \frac{25}{14} ln ∣ x + 7 ∣ + \frac{1}{21} ln ∣ x ∣ + C

求解步骤

\int \frac{x ^{2} + 1}{x ^{3} + 10 x ^{2} + 21 x} d x

乘开

\frac{x ^{2} + 1}{x ^{3} + 10 x ^{2} + 21 x} : \frac{x ^{2}}{x ^{3} + 10 x ^{2} + 21 x} + \frac{1}{x ^{3} + 10 x ^{2} + 21 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{x ^{2}}{x ^{3} + 10 x ^{2} + 21 x} d x + \int \frac{1}{x ^{3} + 10 x ^{2} + 21 x} d x

\int \frac{x ^{2}}{x ^{3} + 10 x ^{2} + 21 x} d x = - \frac{3}{4} ln ∣ x + 3 ∣ + \frac{7}{4} ln ∣ x + 7 ∣

\int \frac{1}{x ^{3} + 10 x ^{2} + 21 x} d x = \frac{1}{21} ln ∣ x ∣ - \frac{1}{12} ln ∣ x + 3 ∣ + \frac{1}{28} ln ∣ x + 7 ∣

= - \frac{3}{4} ln ∣ x + 3 ∣ + \frac{7}{4} ln ∣ x + 7 ∣ + \frac{1}{21} ln ∣ x ∣ - \frac{1}{12} ln ∣ x + 3 ∣ + \frac{1}{28} ln ∣ x + 7 ∣

化简

- \frac{3}{4} ln ∣ x + 3 ∣ + \frac{7}{4} ln ∣ x + 7 ∣ + \frac{1}{21} ln ∣ x ∣ - \frac{1}{12} ln ∣ x + 3 ∣ + \frac{1}{28} ln ∣ x + 7 ∣ : - \frac{5}{6} ln ∣ x + 3 ∣ + \frac{25}{14} ln ∣ x + 7 ∣ + \frac{1}{21} ln ∣ x ∣

= - \frac{5}{6} ln ∣ x + 3 ∣ + \frac{25}{14} ln ∣ x + 7 ∣ + \frac{1}{21} ln ∣ x ∣

解答补常数

= - \frac{5}{6} ln ∣ x + 3 ∣ + \frac{25}{14} ln ∣ x + 7 ∣ + \frac{1}{21} ln ∣ x ∣ + C

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