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

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

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

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

- 2 arcsec (\frac{1}{2} z) + z^{2} - 4 + C

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

= 2 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 2 \cdot \int - 1 + sec^{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

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

\int 1 d u = u

\int sec^{2} (u) d u = tan (u)

= 2 (- u + tan (u))

Substitute back

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

= 2 (- arcsec (\frac{1}{2} z) + tan (arcsec (\frac{1}{2} z)))

Simplify

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

= - 2 arcsec (\frac{1}{2} z) + z^{2} - 4

Add a constant to the solution

= - 2 arcsec (\frac{1}{2} z) + z^{2} - 4 + C

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\int x e^{\frac{3 x ^{2}}{2}} d x

\int 0.024 (0.06 x^{2} - 4 x) + 150 d x

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