integral of t^2sqrt(t+8)

解答

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

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

求解步骤

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

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 = t + 8

代回

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

解答补常数

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

d er i v a t i v e y = \frac{2 x}{x - 1}

x \to 0 lim ((\frac{2 arccos ( x )}{π})^{\frac{1}{x}})

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

d er i v a t i v e f (t) = \frac{sin ( t )}{t}

\int \frac{y}{e ^{- y}} d y