What is the integral of g*m/k (1-e^{-k/m*t}) ?

The integral of g*m/k (1-e^{-k/m*t}) is -(gm^2)/(k^2)(-k/m t-e^{-k/m t})+C

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integral of gm/k (1-e^{-k/mt})

\int g \cdot \frac{m}{k} (1 - e^{- \frac{k}{m} \cdot t}) d t

- \frac{g m ^{2}}{k ^{2}} (- \frac{k}{m} t - e^{- \frac{k}{m} t}) + C

Solution steps

\int g \frac{m}{k} (1 - e^{- \frac{k}{m} t}) d t

Take the constant out:

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

= \frac{m}{k} g \cdot \int 1 - e^{- \frac{k}{m} t} d t

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{g m}{k} \cdot \int 1 - e^{- \frac{k}{m} t} d t

Apply u-substitution

: \int - \frac{m ( 1 - e ^{u} )}{k} d u

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

Take the constant out:

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

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

\int 1 d u = u

\int e^{u} d u = e^{u}

= \frac{g m}{k} (- \frac{m}{k} (u - e^{u}))

Substitute back

u = - \frac{k}{m} t

= \frac{g m}{k} (- \frac{m}{k} (- \frac{k}{m} t - e^{- \frac{k}{m} t}))

Simplify

\frac{g m}{k} (- \frac{m}{k} (- \frac{k}{m} t - e^{- \frac{k}{m} t})) : - \frac{g m ^{2}}{k ^{2}} (- \frac{k}{m} t - e^{- \frac{k}{m} t})

= - \frac{g m ^{2}}{k ^{2}} (- \frac{k}{m} t - e^{- \frac{k}{m} t})

Add a constant to the solution

= - \frac{g m ^{2}}{k ^{2}} (- \frac{k}{m} t - e^{- \frac{k}{m} t}) + C

x \to \frac{5 π}{3} lim (cos (x))

f (x) = x^{3} - x^{2}, at x = - 1

\int x^{2} cos (x^{3} + 1) d x

\int (- \frac{4000}{x ^{2}}) d x

\frac{d}{d x} (3 e^{3 x} cos (2 x) - 2 e^{3 x} sin (2 x))

What is the integral of g*m/k (1-e^{-k/m*t}) ?
The integral of g*m/k (1-e^{-k/m*t}) is -(gm^2)/(k^2)(-k/m t-e^{-k/m t})+C