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受欢迎的微积分 >

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

解答

\int \frac{2}{x ( x ^{2} + 3 ) ^{2}} d x

解答

\frac{2}{9} ln (x) - \frac{1}{9} ln x^{2} + 3 + \frac{1}{3 ( x ^{2} + 3 )} + C

求解步骤

\int \frac{2}{x ( x ^{2} + 3 ) ^{2}} d x

提出常数:

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

= 2 \cdot \int \frac{1}{x ( x ^{2} + 3 ) ^{2}} d x

使用换元积分法

= 2 \cdot \int \frac{1}{2 ( u - 3 ) u ^{2}} d u

提出常数:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{( u - 3 ) u ^{2}} d u

将

\frac{1}{( u - 3 ) u ^{2}}

用部份分式展开:

\frac{1}{9 ( u - 3 )} - \frac{1}{9 u} - \frac{1}{3 u ^{2}}

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{9 ( u - 3 )} - \frac{1}{9 u} - \frac{1}{3 u ^{2}} 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

= 2 \cdot \frac{1}{2} (\int \frac{1}{9 ( u - 3 )} d u - \int \frac{1}{9 u} d u - \int \frac{1}{3 u ^{2}} d u)

\int \frac{1}{9 ( u - 3 )} d u = \frac{1}{9} ln ∣ u - 3 ∣

\int \frac{1}{9 u} d u = \frac{1}{9} ln ∣ u ∣

\int \frac{1}{3 u ^{2}} d u = - \frac{1}{3 u}

= 2 \cdot \frac{1}{2} (\frac{1}{9} ln ∣ u - 3 ∣ - \frac{1}{9} ln ∣ u ∣ - (- \frac{1}{3 u}))

u = x^{2} + 3

代回

= 2 \cdot \frac{1}{2} (\frac{1}{9} ln x^{2} + 3 - 3 - \frac{1}{9} ln x^{2} + 3 - (- \frac{1}{3 ( x ^{2} + 3 )}))

化简

2 \cdot \frac{1}{2} (\frac{1}{9} ln x^{2} + 3 - 3 - \frac{1}{9} ln x^{2} + 3 - (- \frac{1}{3 ( x ^{2} + 3 )})) : \frac{2}{9} ln (x) - \frac{1}{9} ln x^{2} + 3 + \frac{1}{3 ( x ^{2} + 3 )}

= \frac{2}{9} ln (x) - \frac{1}{9} ln x^{2} + 3 + \frac{1}{3 ( x ^{2} + 3 )}

解答补常数

= \frac{2}{9} ln (x) - \frac{1}{9} ln x^{2} + 3 + \frac{1}{3 ( x ^{2} + 3 )} + C

作图

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