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人気のある微分積分 >

integral of arcsin(2x+1)

解

\int arcsin (2 x + 1) d x

解

\frac{1}{2} (arcsin (2 x + 1) (2 x + 1) + 2 - x (x + 1)) + C

解答ステップ

\int arcsin (2 x + 1) d x

u置換積分法を適用する

= \int arcsin (u) \frac{1}{2} d u

定数を除く:

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

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

部分積分を適用する

= \frac{1}{2} (u arcsin (u) - \int \frac{u}{1 - u ^{2}} d u)

\int \frac{u}{1 - u ^{2}} d u = - 1 - u^{2}

= \frac{1}{2} (u arcsin (u) - (- 1 - u^{2}))

代用を戻す

u = 2 x + 1

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

簡素化

\frac{1}{2} ((2 x + 1) arcsin (2 x + 1) - (- 1 - (2 x + 1)^{2})) : \frac{1}{2} (arcsin (2 x + 1) (2 x + 1) + 2 - x (x + 1))

= \frac{1}{2} (arcsin (2 x + 1) (2 x + 1) + 2 - x (x + 1))

定数を解答に追加する

= \frac{1}{2} (arcsin (2 x + 1) (2 x + 1) + 2 - x (x + 1)) + C

グラフ

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