integral of (x+2)/(1-x^4)

解答

\int \frac{x + 2}{1 - x ^{4}} d x

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

求解步骤

\int \frac{x + 2}{1 - x ^{4}} d x

乘开

\frac{x + 2}{1 - x ^{4}} : \frac{x}{1 - x ^{4}} + \frac{2}{1 - x ^{4}}

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}{1 - x ^{4}} d x + \int \frac{2}{1 - x ^{4}} d x

\int \frac{x}{1 - x ^{4}} d x = \frac{1}{4} (ln x^{2} + 1 - ln x^{2} - 1)

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

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

解答补常数

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

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

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