integral of x^2sqrt(5x+4)

解答

\int x^{2} 5 x + 4 d x

\frac{1}{125} (\frac{2}{7} (5 x + 4)^{\frac{7}{2}} - \frac{16}{5} (5 x + 4)^{\frac{5}{2}} + \frac{32}{3} (5 x + 4)^{\frac{3}{2}}) + C

求解步骤

\int x^{2} 5 x + 4 d x

使用换元积分法

= \int \frac{u ( u - 4 ) ^{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 - 4)^{2} d u

乘开

u (u - 4)^{2} : u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 u

= \frac{1}{125} \cdot \int u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 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 8 u^{\frac{3}{2}} d u + \int 16 u d u)

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

\int 8 u^{\frac{3}{2}} d u = \frac{16}{5} u^{\frac{5}{2}}

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

= \frac{1}{125} (\frac{2}{7} u^{\frac{7}{2}} - \frac{16}{5} u^{\frac{5}{2}} + \frac{32}{3} u^{\frac{3}{2}})

u = 5 x + 4

代回

= \frac{1}{125} (\frac{2}{7} (5 x + 4)^{\frac{7}{2}} - \frac{16}{5} (5 x + 4)^{\frac{5}{2}} + \frac{32}{3} (5 x + 4)^{\frac{3}{2}})

解答补常数

= \frac{1}{125} (\frac{2}{7} (5 x + 4)^{\frac{7}{2}} - \frac{16}{5} (5 x + 4)^{\frac{5}{2}} + \frac{32}{3} (5 x + 4)^{\frac{3}{2}}) + C

a re a f (x) = x^{3} - 6 x^{2} + 8, g (x) = x^{2} - 4 x

\frac{d y}{d x} = (x - 8) e^{- 2 y}

d er i v a t i v e f (x) = 2 x + 6

im pl i c i t 2 x^{2} + x y + y^{2} = 4

d er i v a t i v e f (x) = \frac{x ^{2} - 4 x + 3}{x + 1}