integral of x^2sqrt(5x+8)

解答

\int x^{2} 5 x + 8 d x

\frac{1}{125} (\frac{2}{7} (5 x + 8)^{\frac{7}{2}} - \frac{32}{5} (5 x + 8)^{\frac{5}{2}} + \frac{128}{3} (5 x + 8)^{\frac{3}{2}}) + C

求解步骤

\int x^{2} 5 x + 8 d x

使用换元积分法

= \int \frac{u ( u - 8 ) ^{2}}{125} d u

提出常数:

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

= \frac{1}{125} \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}{125} \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}{125} (\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}{125} (\frac{2}{7} u^{\frac{7}{2}} - \frac{32}{5} u^{\frac{5}{2}} + \frac{128}{3} u^{\frac{3}{2}})

u = 5 x + 8

代回

= \frac{1}{125} (\frac{2}{7} (5 x + 8)^{\frac{7}{2}} - \frac{32}{5} (5 x + 8)^{\frac{5}{2}} + \frac{128}{3} (5 x + 8)^{\frac{3}{2}})

解答补常数

= \frac{1}{125} (\frac{2}{7} (5 x + 8)^{\frac{7}{2}} - \frac{32}{5} (5 x + 8)^{\frac{5}{2}} + \frac{128}{3} (5 x + 8)^{\frac{3}{2}}) + C

d er i v a t i v e 6 (x - 3)^{\frac{2}{3}}

\int_{1}^{9} \frac{1}{x - 5} d x

x \to 0 lim (x cos (\frac{1}{x}))

\int x y e^{x + y} d x

\frac{d}{d x} (\frac{( a x + b y ( x ) )}{( c x + d y )})