What is the solution for (dr)/(dt)=((1+e^t)^2)/(e^{3t)} ?

The solution for (dr)/(dt)=((1+e^t)^2)/(e^{3t)} is r=-1/3 e^{-3t}-e^{-2t}-1/(e^t)+c_{1}

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(dr)/(dt)=((1+e^t)^2)/(e^{3t)}

\frac{d r}{d t} = \frac{( 1 + e ^{t} ) ^{2}}{e ^{3 t}}

r = - \frac{1}{3} e^{- 3 t} - e^{- 2 t} - \frac{1}{e ^{t}} + c_{1}

Solution steps

\frac{d r}{d t} = \frac{( 1 + e ^{t} ) ^{2}}{e ^{3 t}}

Substitute

\frac{d r}{d t}

with

r^{'} (t)

r^{^{'}} (t) = \frac{( 1 + e ^{t} ) ^{2}}{e ^{3 t}}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

r = \int \frac{( 1 + e ^{t} ) ^{2}}{e ^{3 t}} d t

\int \frac{( 1 + e ^{t} ) ^{2}}{e ^{3 t}} d t = - \frac{1}{3} e^{- 3 t} - e^{- 2 t} - \frac{1}{e ^{t}} + c_{1}

r = - \frac{1}{3} e^{- 3 t} - e^{- 2 t} - \frac{1}{e ^{t}} + c_{1}

\int x^{2} \cdot e^{- 4 x} d x

\int sec (4 x) tan (4 x) d x

y = \frac{1}{x - 3}

\frac{\partial}{\partial x} (x^{2} y^{3} - 1)

f (x) = (x^{4} + 2 x)^{3}, at x = 2

What is the solution for (dr)/(dt)=((1+e^t)^2)/(e^{3t)} ?
The solution for (dr)/(dt)=((1+e^t)^2)/(e^{3t)} is r=-1/3 e^{-3t}-e^{-2t}-1/(e^t)+c_{1}