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

解答

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

\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2} ln x + 2 + \frac{1}{2} ln x - 2 + C

求解步骤

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

将

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

用部份分式展开:

\frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x - 1 )} - \frac{1}{2 ( x + 2 )} + \frac{1}{2 ( x - 2 )}

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

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

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

\int \frac{1}{2 ( x + 2 )} d x = \frac{1}{2} ln x + 2

\int \frac{1}{2 ( x - 2 )} d x = \frac{1}{2} ln x - 2

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

解答补常数

= \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2} ln x + 2 + \frac{1}{2} ln x - 2 + C

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