What is the integral of 4x-xe^x+e ?

The integral of 4x-xe^x+e is 2x^2-e^xx+e^x+ex+C

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integral of 4x-xe^x+e

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

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

Solution steps

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

\int 4 x d x = 2 x^{2}

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

\int e d x = e x

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

Simplify

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

Add a constant to the solution

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

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

\frac{d}{d x} (- x^{4} cos (x))

n = 1 \sum \infty n^{n}

\int e^{4 x} sin (6 x) d x

\frac{1}{s ( s ( s + 4 ) + 3 )}

What is the integral of 4x-xe^x+e ?
The integral of 4x-xe^x+e is 2x^2-e^xx+e^x+ex+C