What is the integral of xe^{3x}+e^x ?

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

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integral of xe^{3x}+e^x

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

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

Solution steps

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

\int x e^{3 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

f (x) = e^{2 x} (x^{3} + 1)

\frac{d ^{2}}{d x ^{2}} ((4 x^{2} + 9 x) cos (x))

e^{x} y \cdot \frac{d y}{d x} = e^{- y} + e^{- 2 x - y}

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

\frac{d ^{3} y}{d x ^{3}} = 4 x^{3} + x

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