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受欢迎的微积分 >

integral of x/(1-x^8)

解答

\int \frac{x}{1 - x ^{8}} d x

解答

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

求解步骤

\int \frac{x}{1 - x ^{8}} d x

使用换元积分法

= \int \frac{1}{2 ( 1 - u ^{4} )} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{1 - u ^{4}} d u

将

\frac{1}{1 - u ^{4}}

用部份分式展开:

\frac{1}{2 ( u ^{2} + 1 )} + \frac{1}{4 ( u + 1 )} - \frac{1}{4 ( u - 1 )}

= \frac{1}{2} \cdot \int \frac{1}{2 ( u ^{2} + 1 )} + \frac{1}{4 ( u + 1 )} - \frac{1}{4 ( u - 1 )} d u

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

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

\int \frac{1}{2 ( u ^{2} + 1 )} d u = \frac{1}{2} arctan (u)

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

\int \frac{1}{4 ( u - 1 )} d u = \frac{1}{4} ln ∣ u - 1 ∣

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

u = x^{2}

代回

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

解答补常数

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

作图

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