integral of (8x^2+16x)(4x^3+12x^2)^{-3}

解答

\int (8 x^{2} + 16 x) (4 x^{3} + 12 x^{2})^{- 3} d x

- \frac{1}{3 ( 4 x ^{3} + 12 x ^{2} ) ^{2}} + C

求解步骤

\int (8 x^{2} + 16 x) (4 x^{3} + 12 x^{2})^{- 3} d x

使用换元积分法

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

提出常数:

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

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

使用幂法则

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

u = 4 x^{3} + 12 x^{2}

代回

= \frac{2}{3} (- \frac{1}{2 ( 4 x ^{3} + 12 x ^{2} ) ^{2}})

化简

\frac{2}{3} (- \frac{1}{2 ( 4 x ^{3} + 12 x ^{2} ) ^{2}}) : - \frac{1}{3 ( 4 x ^{3} + 12 x ^{2} ) ^{2}}

= - \frac{1}{3 ( 4 x ^{3} + 12 x ^{2} ) ^{2}}

解答补常数

= - \frac{1}{3 ( 4 x ^{3} + 12 x ^{2} ) ^{2}} + C

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