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

The integral of arcsin(1/x) is xarcsin(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 arcsin(1/x)

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

x arcsin (\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 arcsin (\frac{1}{x}) d x

Apply u-substitution

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

Apply Integration By Parts

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

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

t a y l or f (x) = 8.8 x

\frac{d}{d x} (e^{2 x} x (a cos (x) + b sin (x)))

x \to 8 + lim (\frac{3}{x - 8})

\frac{d}{d x} (- \frac{x ^{2}}{2})

d er i v a t i v e 2 5 x^{3}

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