What is the integral of 1/(z^3sqrt(z^2-100)) ?

The integral of 1/(z^3sqrt(z^2-100)) is 1/2000 (arcsec(1/10 z)+1/2 sin(2arcsec(1/10 z)))+C

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integral of 1/(z^3sqrt(z^2-100))

\int \frac{1}{z ^{3} z ^{2} - 100} d z

\frac{1}{2000} (arcsec (\frac{1}{10} z) + \frac{1}{2} sin (2 arcsec (\frac{1}{10} z))) + C

Solution steps

\int \frac{1}{z ^{3} z ^{2} - 100} d z

Apply Trigonometric Substitution

: \int \frac{1}{1000 sec ^{2} ( u )} d u

= \int \frac{1}{1000 sec ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{1000} \cdot \int \frac{1}{sec ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{1000} \cdot \int \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= \frac{1}{1000} \cdot \frac{1}{2} \cdot \int 1 + cos (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

= \frac{1}{1000} \cdot \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= \frac{1}{1000} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsec (\frac{1}{10} z)

= \frac{1}{1000} \cdot \frac{1}{2} (arcsec (\frac{1}{10} z) + \frac{1}{2} sin (2 arcsec (\frac{1}{10} z)))

Simplify

\frac{1}{1000} \cdot \frac{1}{2} (arcsec (\frac{1}{10} z) + \frac{1}{2} sin (2 arcsec (\frac{1}{10} z))) : \frac{1}{2000} (arcsec (\frac{1}{10} z) + \frac{1}{2} sin (2 arcsec (\frac{1}{10} z)))

= \frac{1}{2000} (arcsec (\frac{1}{10} z) + \frac{1}{2} sin (2 arcsec (\frac{1}{10} z)))

Add a constant to the solution

= \frac{1}{2000} (arcsec (\frac{1}{10} z) + \frac{1}{2} sin (2 arcsec (\frac{1}{10} z))) + C

\frac{\partial}{\partial x} (x^{2} y - 3 x y + x y^{2})

\int ln^{2} (x^{2}) d x

\int (2 - \frac{1}{x}) ln (x) d x

\frac{d}{d x} (10 sin (x) + 103 cos (x))

y = 9 x - 6 x, (1, 3)

What is the integral of 1/(z^3sqrt(z^2-100)) ?
The integral of 1/(z^3sqrt(z^2-100)) is 1/2000 (arcsec(1/10 z)+1/2 sin(2arcsec(1/10 z)))+C