What is the integral of g/(1/p-\frac{gz){k}} ?

The integral of g/(1/p-\frac{gz){k}} is -kln| 1/p-(gz)/k |+C

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integral of g/(1/p-\frac{gz){k}}

\int \frac{g}{\frac{1}{p} - \frac{g z}{k}} d z

- k ln \frac{1}{p} - \frac{g z}{k} + C

Solution steps

\int \frac{g}{\frac{1}{p} - \frac{g z}{k}} d z

Take the constant out:

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

= g \cdot \int \frac{1}{\frac{1}{p} - \frac{g z}{k}} d z

Apply u-substitution

: \int - \frac{k}{gu} d u

= g \cdot \int - \frac{k}{gu} d u

Take the constant out:

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

= g (- \frac{k}{g} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= g (- \frac{k}{g} ln ∣ u ∣)

Substitute back

u = \frac{1}{p} - \frac{g z}{k}

= g (- \frac{k}{g} ln \frac{1}{p} - \frac{g z}{k})

Simplify

g (- \frac{k}{g} ln \frac{1}{p} - \frac{g z}{k}) : - k ln \frac{1}{p} - \frac{g z}{k}

= - k ln \frac{1}{p} - \frac{g z}{k}

Add a constant to the solution

= - k ln \frac{1}{p} - \frac{g z}{k} + C

\int_{0}^{2} 0.2 e^{- 0.2 x} d x

y^{^{'}} - 5 y = t e^{(- 4 t)}

f (x) = \frac{1}{x ^{2}}, at x = 2

x y^{^{'}} = y + x^{2} sin (x), y (\frac{π}{3}) = 0

h (31 - \frac{1}{2} h)^{2}

What is the integral of g/(1/p-\frac{gz){k}} ?
The integral of g/(1/p-\frac{gz){k}} is -kln| 1/p-(gz)/k |+C