What is the integral of 1/(z^3sqrt(z^2-49)) ?

The integral of 1/(z^3sqrt(z^2-49)) is 1/686 (arcsec(1/7 z)+1/2 sin(2arcsec(1/7 z)))+C

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integral of 1/(z^3sqrt(z^2-49))

\int \frac{1}{z ^{3} z ^{2} - 49} d z

\frac{1}{686} (arcsec (\frac{1}{7} z) + \frac{1}{2} sin (2 arcsec (\frac{1}{7} z))) + C

Solution steps

\int \frac{1}{z ^{3} z ^{2} - 49} d z

Apply Trigonometric Substitution

: \int \frac{1}{343 sec ^{2} ( u )} d u

= \int \frac{1}{343 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}{343} \cdot \int \frac{1}{sec ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{343} \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}{343} \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}{343} \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}{343} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsec (\frac{1}{7} z)

= \frac{1}{343} \cdot \frac{1}{2} (arcsec (\frac{1}{7} z) + \frac{1}{2} sin (2 arcsec (\frac{1}{7} z)))

Simplify

\frac{1}{343} \cdot \frac{1}{2} (arcsec (\frac{1}{7} z) + \frac{1}{2} sin (2 arcsec (\frac{1}{7} z))) : \frac{1}{686} (arcsec (\frac{1}{7} z) + \frac{1}{2} sin (2 arcsec (\frac{1}{7} z)))

= \frac{1}{686} (arcsec (\frac{1}{7} z) + \frac{1}{2} sin (2 arcsec (\frac{1}{7} z)))

Add a constant to the solution

= \frac{1}{686} (arcsec (\frac{1}{7} z) + \frac{1}{2} sin (2 arcsec (\frac{1}{7} z))) + C

\frac{d}{d x} ((x + 4)^{\frac{2}{3}})

\int \frac{1}{- y ^{2} + y + 20} d y

x \to 2 lim (\frac{- 1}{0})

y = (1 + 5 x)^{10}, (0, 1)

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

What is the integral of 1/(z^3sqrt(z^2-49)) ?
The integral of 1/(z^3sqrt(z^2-49)) is 1/686 (arcsec(1/7 z)+1/2 sin(2arcsec(1/7 z)))+C