What is the integral of e^{3x}(x+e^{-2x}) ?

The integral of e^{3x}(x+e^{-2x}) is 1/9 (e^{3x}*3x-e^{3x})+e^x+C

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integral of e^{3x}(x+e^{-2x})

\int e^{3 x} (x + e^{- 2 x}) d x

\frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) + e^{x} + C

Solution steps

\int e^{3 x} (x + e^{- 2 x}) d x

Expand

e^{3 x} (x + e^{- 2 x}) : e^{3 x} x + e^{x}

= \int e^{3 x} 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 e^{3 x} x d x + \int e^{x} d x

\int e^{3 x} x d x = \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x})

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

= \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) + e^{x}

Add a constant to the solution

= \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) + e^{x} + C

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What is the integral of e^{3x}(x+e^{-2x}) ?
The integral of e^{3x}(x+e^{-2x}) is 1/9 (e^{3x}*3x-e^{3x})+e^x+C