What is the integral of arccosh(x) ?

The integral of arccosh(x) is xarccosh(x)-sqrt((x-1)(x+1))+C

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integral of arccosh(x)

\int arccosh (x) d x

x arccosh (x) - (x - 1) (x + 1) + C

Solution steps

\int arccosh (x) d x

Apply Integration By Parts

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

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

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

= x arccosh (x) - (x - 1) (x + 1)

Add a constant to the solution

= x arccosh (x) - (x - 1) (x + 1) + C

8 y^{^{'}} + y = 6 t^{2}

\int 1 + 3 x d x

y = cos (x^{2}) 2 x

n = 1 \sum \infty \frac{( e ^{\frac{8}{n}} )}{n}

\frac{d}{d x} ((\frac{x - 7}{x - 9})^{4})

What is the integral of arccosh(x) ?
The integral of arccosh(x) is xarccosh(x)-sqrt((x-1)(x+1))+C