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受欢迎的微积分 >

integral of (-8)/((x-2)(x^2+4))

解答

\int \frac{- 8}{( x - 2 ) ( x ^{2} + 4 )} d x

解答

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

求解步骤

\int \frac{- 8}{( x - 2 ) ( x ^{2} + 4 )} d x

提出常数:

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

= - 8 \cdot \int \frac{1}{( x - 2 ) ( x ^{2} + 4 )} d x

使用换元积分法

= - 8 \cdot \int \frac{1}{4 ( u - 1 ) ( u ^{2} + 1 )} d u

提出常数:

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

= - 8 \cdot \frac{1}{4} \cdot \int \frac{1}{( u - 1 ) ( u ^{2} + 1 )} d u

将

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

用部份分式展开:

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

= - 8 \cdot \frac{1}{4} \cdot \int \frac{1}{2 ( u - 1 )} + \frac{- u - 1}{2 ( u ^{2} + 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

= - 8 \cdot \frac{1}{4} (\int \frac{1}{2 ( u - 1 )} d u + \int \frac{- u - 1}{2 ( u ^{2} + 1 )} d u)

\int \frac{1}{2 ( u - 1 )} d u = \frac{1}{2} ln ∣ u - 1 ∣

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

= - 8 \cdot \frac{1}{4} (\frac{1}{2} ln ∣ u - 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln u^{2} + 1 - arctan (u)))

u = \frac{x}{2}

代回

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

化简

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

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

解答补常数

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

作图

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