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

The integral of (sqrt(1/z-x^2))/(x^2) is -arcsin(sqrt(z)x)-(sqrt(1-zx^2))/(sqrt(z)x)+C

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int cot^{2} (u) d u

= \int cot^{2} (u) d u

Rewrite using trig identities

= \int - 1 + csc^{2} (u) 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 1 d u + \int csc^{2} (u) d u

\int 1 d u = u

\int csc^{2} (u) d u = - cot (u)

= - u - cot (u)

Substitute back

u = arcsin \frac{1}{\frac{1}{z}} x

= - arcsin \frac{1}{\frac{1}{z}} x - cot arcsin \frac{1}{\frac{1}{z}} x

Simplify

- arcsin \frac{1}{\frac{1}{z}} x - cot arcsin \frac{1}{\frac{1}{z}} x : - arcsin (z x) - \frac{1 - z x ^{2}}{z x}

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

Add a constant to the solution

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

\int (sec^{2} (θ) - sin (θ)) d θ

\int - 3 t^{2} cos (t) d t

n = 1 \sum \infty \frac{- 3}{2 ^{n}}

x, x^{2}, - x + 1

\frac{d}{d x} (ln (\frac{x - 1}{x + 1}))

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