integral of xsqrt(3x+8)

解答

\int x 3 x + 8 d x

\frac{1}{9} (\frac{2}{5} (3 x + 8)^{\frac{5}{2}} - \frac{16}{3} (3 x + 8)^{\frac{3}{2}}) + C

求解步骤

\int x 3 x + 8 d x

使用换元积分法

= \int \frac{u ( u - 8 )}{9} d u

提出常数:

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

= \frac{1}{9} \cdot \int u (u - 8) d u

乘开

u (u - 8) : u^{\frac{3}{2}} - 8 u

= \frac{1}{9} \cdot \int u^{\frac{3}{2}} - 8 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}{9} (\int u^{\frac{3}{2}} d u - \int 8 u d u)

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

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

= \frac{1}{9} (\frac{2}{5} u^{\frac{5}{2}} - \frac{16}{3} u^{\frac{3}{2}})

u = 3 x + 8

代回

= \frac{1}{9} (\frac{2}{5} (3 x + 8)^{\frac{5}{2}} - \frac{16}{3} (3 x + 8)^{\frac{3}{2}})

解答补常数

= \frac{1}{9} (\frac{2}{5} (3 x + 8)^{\frac{5}{2}} - \frac{16}{3} (3 x + 8)^{\frac{3}{2}}) + C

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