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受欢迎的微积分 >

integral of (16x^3)/(4x^2-4x+1)

解答

\int \frac{16 x ^{3}}{4 x ^{2} - 4 x + 1} d x

解答

\frac{8 x ^{3} + 12 x ^{2} + 12 x ln x - \frac{1}{2} - 18 x - 6 ln x - \frac{1}{2} + 3}{2 ( 2 x - 1 )} + C

求解步骤

\int \frac{16 x ^{3}}{4 x ^{2} - 4 x + 1} d x

提出常数:

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

= 16 \cdot \int \frac{x ^{3}}{4 x ^{2} - 4 x + 1} d x

对

4 x^{2} - 4 x + 1

配方:

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

= 16 \cdot \int \frac{x ^{3}}{4 ( x - \frac{1}{2} ) ^{2}} d x

使用换元积分法

= 16 \cdot \int \frac{( 2 u + 1 ) ^{3}}{32 u ^{2}} d u

提出常数:

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

= 16 \cdot \frac{1}{32} \cdot \int \frac{( 2 u + 1 ) ^{3}}{u ^{2}} d u

乘开

\frac{( 2 u + 1 ) ^{3}}{u ^{2}} : 8 u + 12 + \frac{6}{u} + \frac{1}{u ^{2}}

= 16 \cdot \frac{1}{32} \cdot \int 8 u + 12 + \frac{6}{u} + \frac{1}{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

= 16 \cdot \frac{1}{32} (\int 8 u d u + \int 12 d u + \int \frac{6}{u} d u + \int \frac{1}{u ^{2}} d u)

\int 8 u d u = 4 u^{2}

\int 12 d u = 12 u

\int \frac{6}{u} d u = 6 ln ∣ u ∣

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

= 16 \cdot \frac{1}{32} (4 u^{2} + 12 u + 6 ln ∣ u ∣ - \frac{1}{u})

u = x - \frac{1}{2}

代回

= 16 \cdot \frac{1}{32} (4 (x - \frac{1}{2})^{2} + 12 (x - \frac{1}{2}) + 6 ln x - \frac{1}{2} - \frac{1}{x - \frac{1}{2}})

化简

16 \cdot \frac{1}{32} (4 (x - \frac{1}{2})^{2} + 12 (x - \frac{1}{2}) + 6 ln x - \frac{1}{2} - \frac{1}{x - \frac{1}{2}}) : \frac{8 x ^{3} + 12 x ^{2} + 12 x ln x - \frac{1}{2} - 18 x - 6 ln x - \frac{1}{2} + 3}{2 ( 2 x - 1 )}

= \frac{8 x ^{3} + 12 x ^{2} + 12 x ln x - \frac{1}{2} - 18 x - 6 ln x - \frac{1}{2} + 3}{2 ( 2 x - 1 )}

解答补常数

= \frac{8 x ^{3} + 12 x ^{2} + 12 x ln x - \frac{1}{2} - 18 x - 6 ln x - \frac{1}{2} + 3}{2 ( 2 x - 1 )} + C

作图

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