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人気のある微分積分 >

integral of arccosh(2x)

解

\int arccosh (2 x) d x

解

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

解答ステップ

\int arccosh (2 x) d x

u置換積分法を適用する

= \int arccosh (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 arccosh (u) d u

部分積分を適用する

= \frac{1}{2} (u arccosh (u) - \int \frac{u}{u - 1 u + 1} d u)

\int \frac{u}{u - 1 u + 1} d u = (u - 1) (u + 1)

= \frac{1}{2} (u arccosh (u) - (u - 1) (u + 1))

代用を戻す

u = 2 x

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

簡素化

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

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

定数を解答に追加する

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

グラフ

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