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受欢迎的微积分 >

integral of 3/(x(x^2+6)^2)

解答

\int \frac{3}{x ( x ^{2} + 6 ) ^{2}} d x

解答

\frac{3}{2} (\frac{1}{18} ln (x) - \frac{1}{36} ln x^{2} + 6 + \frac{1}{6 ( x ^{2} + 6 )}) + C

求解步骤

\int \frac{3}{x ( x ^{2} + 6 ) ^{2}} d x

提出常数:

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

= 3 \cdot \int \frac{1}{x ( x ^{2} + 6 ) ^{2}} d x

使用换元积分法

= 3 \cdot \int \frac{1}{2 ( u - 6 ) u ^{2}} d u

提出常数:

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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{( u - 6 ) u ^{2}} d u

将

\frac{1}{( u - 6 ) u ^{2}}

用部份分式展开:

\frac{1}{36 ( u - 6 )} - \frac{1}{36 u} - \frac{1}{6 u ^{2}}

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{36 ( u - 6 )} - \frac{1}{36 u} - \frac{1}{6 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

= 3 \cdot \frac{1}{2} (\int \frac{1}{36 ( u - 6 )} d u - \int \frac{1}{36 u} d u - \int \frac{1}{6 u ^{2}} d u)

\int \frac{1}{36 ( u - 6 )} d u = \frac{1}{36} ln ∣ u - 6 ∣

\int \frac{1}{36 u} d u = \frac{1}{36} ln ∣ u ∣

\int \frac{1}{6 u ^{2}} d u = - \frac{1}{6 u}

= 3 \cdot \frac{1}{2} (\frac{1}{36} ln ∣ u - 6 ∣ - \frac{1}{36} ln ∣ u ∣ - (- \frac{1}{6 u}))

u = x^{2} + 6

代回

= 3 \cdot \frac{1}{2} (\frac{1}{36} ln x^{2} + 6 - 6 - \frac{1}{36} ln x^{2} + 6 - (- \frac{1}{6 ( x ^{2} + 6 )}))

化简

3 \cdot \frac{1}{2} (\frac{1}{36} ln x^{2} + 6 - 6 - \frac{1}{36} ln x^{2} + 6 - (- \frac{1}{6 ( x ^{2} + 6 )})) : \frac{3}{2} (\frac{1}{18} ln (x) - \frac{1}{36} ln x^{2} + 6 + \frac{1}{6 ( x ^{2} + 6 )})

= \frac{3}{2} (\frac{1}{18} ln (x) - \frac{1}{36} ln x^{2} + 6 + \frac{1}{6 ( x ^{2} + 6 )})

解答补常数

= \frac{3}{2} (\frac{1}{18} ln (x) - \frac{1}{36} ln x^{2} + 6 + \frac{1}{6 ( x ^{2} + 6 )}) + C

作图

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