integral of x^2sqrt(7x+8)

解答

\int x^{2} 7 x + 8 d x

\frac{1}{343} (\frac{2}{7} (7 x + 8)^{\frac{7}{2}} - \frac{32}{5} (7 x + 8)^{\frac{5}{2}} + \frac{128}{3} (7 x + 8)^{\frac{3}{2}}) + C

求解步骤

\int x^{2} 7 x + 8 d x

使用换元积分法

= \int \frac{u ( u - 8 ) ^{2}}{343} d u

提出常数:

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

= \frac{1}{343} \cdot \int u (u - 8)^{2} d u

乘开

u (u - 8)^{2} : u^{\frac{5}{2}} - 16 u^{\frac{3}{2}} + 64 u

= \frac{1}{343} \cdot \int u^{\frac{5}{2}} - 16 u^{\frac{3}{2}} + 64 u 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

= \frac{1}{343} (\int u^{\frac{5}{2}} d u - \int 16 u^{\frac{3}{2}} d u + \int 64 u d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 16 u^{\frac{3}{2}} d u = \frac{32}{5} u^{\frac{5}{2}}

\int 64 u d u = \frac{128}{3} u^{\frac{3}{2}}

= \frac{1}{343} (\frac{2}{7} u^{\frac{7}{2}} - \frac{32}{5} u^{\frac{5}{2}} + \frac{128}{3} u^{\frac{3}{2}})

u = 7 x + 8

代回

= \frac{1}{343} (\frac{2}{7} (7 x + 8)^{\frac{7}{2}} - \frac{32}{5} (7 x + 8)^{\frac{5}{2}} + \frac{128}{3} (7 x + 8)^{\frac{3}{2}})

解答补常数

= \frac{1}{343} (\frac{2}{7} (7 x + 8)^{\frac{7}{2}} - \frac{32}{5} (7 x + 8)^{\frac{5}{2}} + \frac{128}{3} (7 x + 8)^{\frac{3}{2}}) + C

s im pl i f y \frac{d}{dd} (a x^{3} + b x^{2} + c x + d)

\frac{d}{d x} ((x^{3} - 2 x) (- x + 6 x^{2}))

(e^{2 x} + 9) \frac{d y}{d x} = y

\int (x^{2} + 2 x + 1) d x

d er i v a t i v e f (x) = (- 25 x^{2} + 1)^{2}