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受欢迎的微积分 >

integral of 4x(3x-1)^5

解答

\int 4 x (3 x - 1)^{5} d x

解答

\frac{2}{189} (6 (3 x - 1)^{7} + 7 (3 x - 1)^{6}) + C

求解步骤

\int 4 x (3 x - 1)^{5} d x

提出常数:

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

= 4 \cdot \int x (3 x - 1)^{5} d x

使用换元积分法

= 4 \cdot \int \frac{u ^{5} ( u + 1 )}{9} d u

提出常数:

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

= 4 \cdot \frac{1}{9} \cdot \int u^{5} (u + 1) d u

乘开

u^{5} (u + 1) : u^{6} + u^{5}

= 4 \cdot \frac{1}{9} \cdot \int u^{6} + u^{5} 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

= 4 \cdot \frac{1}{9} (\int u^{6} d u + \int u^{5} d u)

\int u^{6} d u = \frac{u ^{7}}{7}

\int u^{5} d u = \frac{u ^{6}}{6}

= 4 \cdot \frac{1}{9} (\frac{u ^{7}}{7} + \frac{u ^{6}}{6})

u = 3 x - 1

代回

= 4 \cdot \frac{1}{9} (\frac{( 3 x - 1 ) ^{7}}{7} + \frac{( 3 x - 1 ) ^{6}}{6})

化简

4 \cdot \frac{1}{9} (\frac{( 3 x - 1 ) ^{7}}{7} + \frac{( 3 x - 1 ) ^{6}}{6}) : \frac{2}{189} (6 (3 x - 1)^{7} + 7 (3 x - 1)^{6})

= \frac{2}{189} (6 (3 x - 1)^{7} + 7 (3 x - 1)^{6})

解答补常数

= \frac{2}{189} (6 (3 x - 1)^{7} + 7 (3 x - 1)^{6}) + C

作图

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