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受欢迎的微积分 >

integral of 5/(x(x^2+4)^2)

解答

\int \frac{5}{x ( x ^{2} + 4 ) ^{2}} d x

解答

\frac{5}{2} (\frac{1}{8} ln (x) - \frac{1}{16} ln x^{2} + 4 + \frac{1}{4 ( x ^{2} + 4 )}) + C

求解步骤

\int \frac{5}{x ( x ^{2} + 4 ) ^{2}} d x

提出常数:

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

= 5 \cdot \int \frac{1}{x ( x ^{2} + 4 ) ^{2}} d x

使用换元积分法

= 5 \cdot \int \frac{1}{2 ( u - 4 ) u ^{2}} d u

提出常数:

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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{( u - 4 ) u ^{2}} d u

将

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

用部份分式展开:

\frac{1}{16 ( u - 4 )} - \frac{1}{16 u} - \frac{1}{4 u ^{2}}

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{16 ( u - 4 )} - \frac{1}{16 u} - \frac{1}{4 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

= 5 \cdot \frac{1}{2} (\int \frac{1}{16 ( u - 4 )} d u - \int \frac{1}{16 u} d u - \int \frac{1}{4 u ^{2}} d u)

\int \frac{1}{16 ( u - 4 )} d u = \frac{1}{16} ln ∣ u - 4 ∣

\int \frac{1}{16 u} d u = \frac{1}{16} ln ∣ u ∣

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

= 5 \cdot \frac{1}{2} (\frac{1}{16} ln ∣ u - 4 ∣ - \frac{1}{16} ln ∣ u ∣ - (- \frac{1}{4 u}))

u = x^{2} + 4

代回

= 5 \cdot \frac{1}{2} (\frac{1}{16} ln x^{2} + 4 - 4 - \frac{1}{16} ln x^{2} + 4 - (- \frac{1}{4 ( x ^{2} + 4 )}))

化简

5 \cdot \frac{1}{2} (\frac{1}{16} ln x^{2} + 4 - 4 - \frac{1}{16} ln x^{2} + 4 - (- \frac{1}{4 ( x ^{2} + 4 )})) : \frac{5}{2} (\frac{1}{8} ln (x) - \frac{1}{16} ln x^{2} + 4 + \frac{1}{4 ( x ^{2} + 4 )})

= \frac{5}{2} (\frac{1}{8} ln (x) - \frac{1}{16} ln x^{2} + 4 + \frac{1}{4 ( x ^{2} + 4 )})

解答补常数

= \frac{5}{2} (\frac{1}{8} ln (x) - \frac{1}{16} ln x^{2} + 4 + \frac{1}{4 ( x ^{2} + 4 )}) + C

作图

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