What is the integral of (4xe^x+e^{4x})/(e^{3x)} ?

The integral of (4xe^x+e^{4x})/(e^{3x)} is -2e^{-2x}x-e^{-2x}+e^x+C

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integral of (4xe^x+e^{4x})/(e^{3x)}

\int \frac{4 x e ^{x} + e ^{4 x}}{e ^{3 x}} d x

- 2 e^{- 2 x} x - e^{- 2 x} + e^{x} + C

Solution steps

\int \frac{4 x e ^{x} + e ^{4 x}}{e ^{3 x}} d x

Expand

\frac{4 x e ^{x} + e ^{4 x}}{e ^{3 x}} : \frac{4 x}{e ^{2 x}} + e^{x}

= \int \frac{4 x}{e ^{2 x}} + e^{x} d x

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{4 x}{e ^{2 x}} d x + \int e^{x} d x

\int \frac{4 x}{e ^{2 x}} d x = - 2 e^{- 2 x} x - e^{- 2 x}

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

= - 2 e^{- 2 x} x - e^{- 2 x} + e^{x}

Add a constant to the solution

= - 2 e^{- 2 x} x - e^{- 2 x} + e^{x} + C

\int_{1}^{4} f (x) d x

\frac{d}{d x} (x (x^{5} + 1)^{3})

\frac{\partial}{\partial y} (x e^{y} - ln (x))

- x - \frac{1}{x}

x \to 0 lim (x^{- 1})

What is the integral of (4xe^x+e^{4x})/(e^{3x)} ?
The integral of (4xe^x+e^{4x})/(e^{3x)} is -2e^{-2x}x-e^{-2x}+e^x+C