integral of x^2sqrt(8+x)

解答

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

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

求解步骤

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

使用换元积分法

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

乘开

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

= \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

= \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{2}{7} u^{\frac{7}{2}} - \frac{32}{5} u^{\frac{5}{2}} + \frac{128}{3} u^{\frac{3}{2}}

u = 8 + x

代回

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

解答补常数

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

\int \frac{7}{9 + 4 x ^{2}} d x

\frac{d}{d x} (ln (\frac{x}{7}))

y^{^{''}} + 5 y^{^{'}} + 6 = 1

l a pl a ce t r an s f or m 3 e^{- 4 t}

d er i v a t i v e (4 x + 3) (2 x^{2} + 7 x - 1)