What is the integral of 1/(z^2sqrt(z^2-4)) ?

The integral of 1/(z^2sqrt(z^2-4)) is (sqrt(z^2-4))/(4z)+C

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integral of 1/(z^2sqrt(z^2-4))

\int \frac{1}{z ^{2} z ^{2} - 4} d z

\frac{z ^{2} - 4}{4 z} + C

Solution steps

\int \frac{1}{z ^{2} z ^{2} - 4} d z

Apply Trigonometric Substitution

: \int \frac{1}{4 sec ( u )} d u

= \int \frac{1}{4 sec ( u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{4} sin (u)

Substitute back

u = arcsec (\frac{1}{2} z)

= \frac{1}{4} sin (arcsec (\frac{1}{2} z))

Simplify

\frac{1}{4} sin (arcsec (\frac{1}{2} z)) : \frac{z ^{2} - 4}{4 z}

= \frac{z ^{2} - 4}{4 z}

Add a constant to the solution

= \frac{z ^{2} - 4}{4 z} + C

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What is the integral of 1/(z^2sqrt(z^2-4)) ?
The integral of 1/(z^2sqrt(z^2-4)) is (sqrt(z^2-4))/(4z)+C