What is the integral of sqrt(1-1/(x^2)) ?

The integral of sqrt(1-1/(x^2)) is -arctan(sqrt(x^2-1))+sqrt(x^2-1)+C

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integral of sqrt(1-1/(x^2))

\int 1 - \frac{1}{x ^{2}} d x

- arctan (x^{2} - 1) + x^{2} - 1 + C

Solution steps

\int 1 - \frac{1}{x ^{2}} d x

Simplify

1 - \frac{1}{x ^{2}} : \frac{x ^{2} - 1}{x}

= \int \frac{x ^{2} - 1}{x} d x

Apply u-substitution

: \int \frac{u ^{2}}{u ^{2} + 1} d u

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

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int 1 d u = u

= - arctan (u) + u

Substitute back

u = x^{2} - 1

= - arctan (x^{2} - 1) + x^{2} - 1

Add a constant to the solution

= - arctan (x^{2} - 1) + x^{2} - 1 + C

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What is the integral of sqrt(1-1/(x^2)) ?
The integral of sqrt(1-1/(x^2)) is -arctan(sqrt(x^2-1))+sqrt(x^2-1)+C