What is the integral of arccos(1/x) ?

The integral of arccos(1/x) is xarccos(1/x)-1/2 ln|(sqrt(x^2-1))/x+1|+1/2 ln|(sqrt(x^2-1))/x-1|+C

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integral of arccos(1/x)

\int arccos (\frac{1}{x}) d x

x arccos (\frac{1}{x}) - \frac{1}{2} ln \frac{x ^{2} - 1}{x} + 1 + \frac{1}{2} ln \frac{x ^{2} - 1}{x} - 1 + C

Solution steps

\int arccos (\frac{1}{x}) d x

Apply u-substitution

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

= - (- \frac{arccos ( u )}{u} - (- \frac{1}{2} ln 1 - u^{2} + 1 + \frac{1}{2} ln 1 - u^{2} - 1))

Substitute back

u = \frac{1}{x}

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

Simplify

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

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

Add a constant to the solution

= x arccos (\frac{1}{x}) - \frac{1}{2} ln \frac{x ^{2} - 1}{x} + 1 + \frac{1}{2} ln \frac{x ^{2} - 1}{x} - 1 + C

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What is the integral of arccos(1/x) ?
The integral of arccos(1/x) is xarccos(1/x)-1/2 ln|(sqrt(x^2-1))/x+1|+1/2 ln|(sqrt(x^2-1))/x-1|+C