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

解答

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

\frac{1}{2} (u + 2)^{2} - 6 (u + 2) + 12 ln ∣ u + 2 ∣ + \frac{8}{u + 2} + C

求解步骤

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

使用换元积分法

= \int \frac{( v - 2 ) ^{3}}{v ^{2}} d v

乘开

\frac{( v - 2 ) ^{3}}{v ^{2}} : v - 6 + \frac{12}{v} - \frac{8}{v ^{2}}

= \int v - 6 + \frac{12}{v} - \frac{8}{v ^{2}} d v

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

= \int v d v - \int 6 d v + \int \frac{12}{v} d v - \int \frac{8}{v ^{2}} d v

\int v d v = \frac{v ^{2}}{2}

\int 6 d v = 6 v

\int \frac{12}{v} d v = 12 ln ∣ v ∣

\int \frac{8}{v ^{2}} d v = - \frac{8}{v}

= \frac{v ^{2}}{2} - 6 v + 12 ln ∣ v ∣ - (- \frac{8}{v})

v = u + 2

代回

= \frac{( u + 2 ) ^{2}}{2} - 6 (u + 2) + 12 ln ∣ u + 2 ∣ - (- \frac{8}{u + 2})

化简

\frac{( u + 2 ) ^{2}}{2} - 6 (u + 2) + 12 ln ∣ u + 2 ∣ - (- \frac{8}{u + 2}) : \frac{1}{2} (u + 2)^{2} - 6 (u + 2) + 12 ln ∣ u + 2 ∣ + \frac{8}{u + 2}

= \frac{1}{2} (u + 2)^{2} - 6 (u + 2) + 12 ln ∣ u + 2 ∣ + \frac{8}{u + 2}

解答补常数

= \frac{1}{2} (u + 2)^{2} - 6 (u + 2) + 12 ln ∣ u + 2 ∣ + \frac{8}{u + 2} + C

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