integral of sqrt((2x+1)^5)

解答

\int (2 x + 1)^{5} d x

(2 x + 1)^{\frac{5}{2}} x - 5 (\frac{2}{7} x^{2} (2 x + 1)^{\frac{3}{2}} + \frac{3}{35} x (2 x + 1)^{\frac{3}{2}} - \frac{1}{35} (2 x + 1)^{\frac{3}{2}}) + C

求解步骤

\int (2 x + 1)^{5} d x

(2 x + 1)^{5} = (2 x + 1)^{\frac{5}{2}},

假定

(2 x + 1) \geq 0

= \int (2 x + 1)^{\frac{5}{2}} d x

使用分布积分法

= (2 x + 1)^{\frac{5}{2}} x - \int 5 (2 x + 1)^{\frac{3}{2}} x d x

\int 5 (2 x + 1)^{\frac{3}{2}} x d x = 5 (\frac{2}{7} x^{2} (2 x + 1)^{\frac{3}{2}} + \frac{3}{35} x (2 x + 1)^{\frac{3}{2}} - \frac{1}{35} (2 x + 1)^{\frac{3}{2}})

= (2 x + 1)^{\frac{5}{2}} x - 5 (\frac{2}{7} x^{2} (2 x + 1)^{\frac{3}{2}} + \frac{3}{35} x (2 x + 1)^{\frac{3}{2}} - \frac{1}{35} (2 x + 1)^{\frac{3}{2}})

解答补常数

= (2 x + 1)^{\frac{5}{2}} x - 5 (\frac{2}{7} x^{2} (2 x + 1)^{\frac{3}{2}} + \frac{3}{35} x (2 x + 1)^{\frac{3}{2}} - \frac{1}{35} (2 x + 1)^{\frac{3}{2}}) + C

in v erse l a pl a ce \frac{- 2}{( s ^{2} + 1 ) ^{2}}

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