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人気のある微分積分 >

integral of 2xsqrt(2x-1)

解

\int 2 x 2 x - 1 d x

解

\frac{2}{15} (3 x + 1) (2 x - 1)^{\frac{3}{2}} + C

解答ステップ

\int 2 x 2 x - 1 d x

定数を除く:

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

= 2 \cdot \int x 2 x - 1 d x

u置換積分法を適用する

= 2 \cdot \int \frac{u ( u + 1 )}{4} d u

定数を除く:

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

= 2 \cdot \frac{1}{4} \cdot \int u (u + 1) d u

拡張

u (u + 1) : u^{\frac{3}{2}} + u

= 2 \cdot \frac{1}{4} \cdot \int u^{\frac{3}{2}} + u d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{4} (\int u^{\frac{3}{2}} d u + \int u d u)

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

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

= 2 \cdot \frac{1}{4} (\frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

代用を戻す

u = 2 x - 1

= 2 \cdot \frac{1}{4} (\frac{2}{5} (2 x - 1)^{\frac{5}{2}} + \frac{2}{3} (2 x - 1)^{\frac{3}{2}})

簡素化

2 \cdot \frac{1}{4} (\frac{2}{5} (2 x - 1)^{\frac{5}{2}} + \frac{2}{3} (2 x - 1)^{\frac{3}{2}}) : \frac{2}{15} (3 x + 1) (2 x - 1)^{\frac{3}{2}}

= \frac{2}{15} (3 x + 1) (2 x - 1)^{\frac{3}{2}}

定数を解答に追加する

= \frac{2}{15} (3 x + 1) (2 x - 1)^{\frac{3}{2}} + C

グラフ

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