integral of arccosh(2x)

解答

\int arccosh (2 x) d x

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

求解步骤

\int arccosh (2 x) d x

使用换元积分法

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

提出常数:

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

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

使用分布积分法

= \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))

u = 2 x

代回

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

化简

\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)

解答补常数

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

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