What is the integral of (e^t+2e^{2t})/(e^{3t)} ?

The integral of (e^t+2e^{2t})/(e^{3t)} is -1/2 e^{-2t}-2/(e^t)+C

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integral of (e^t+2e^{2t})/(e^{3t)}

\int \frac{e ^{t} + 2 e ^{2 t}}{e ^{3 t}} d t

- \frac{1}{2} e^{- 2 t} - \frac{2}{e ^{t}} + C

Solution steps

\int \frac{e ^{t} + 2 e ^{2 t}}{e ^{3 t}} d t

Expand

\frac{e ^{t} + 2 e ^{2 t}}{e ^{3 t}} : \frac{1}{e ^{2 t}} + \frac{2}{e ^{t}}

= \int \frac{1}{e ^{2 t}} + \frac{2}{e ^{t}} d t

Apply the Sum Rule:

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

= \int \frac{1}{e ^{2 t}} d t + \int \frac{2}{e ^{t}} d t

\int \frac{1}{e ^{2 t}} d t = - \frac{1}{2} e^{- 2 t}

\int \frac{2}{e ^{t}} d t = - \frac{2}{e ^{t}}

= - \frac{1}{2} e^{- 2 t} - \frac{2}{e ^{t}}

Add a constant to the solution

= - \frac{1}{2} e^{- 2 t} - \frac{2}{e ^{t}} + C

\frac{\partial}{\partial x} (x e^{5 y})

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

\frac{\partial}{\partial y} (2 x^{2} y)

\int_{1}^{2} x sin (π x) d x

\frac{d y}{d t} + 2 y = e^{- t} cos (3 t), y (0) = 0

What is the integral of (e^t+2e^{2t})/(e^{3t)} ?
The integral of (e^t+2e^{2t})/(e^{3t)} is -1/2 e^{-2t}-2/(e^t)+C