What is the integral of (4x^2-x)e^x ?

The integral of (4x^2-x)e^x is 4e^xx^2-9e^xx+9e^x+C

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integral of (4x^2-x)e^x

\int (4 x^{2} - x) e^{x} d x

4 e^{x} x^{2} - 9 e^{x} x + 9 e^{x} + C

Solution steps

\int (4 x^{2} - x) e^{x} d x

Expand

(4 x^{2} - x) e^{x} : 4 e^{x} x^{2} - e^{x} x

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

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

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

= 4 (x^{2} e^{x} - 2 (e^{x} x - e^{x})) - (e^{x} x - e^{x})

Simplify

4 (x^{2} e^{x} - 2 (e^{x} x - e^{x})) - (e^{x} x - e^{x}) : 4 e^{x} x^{2} - 9 e^{x} x + 9 e^{x}

= 4 e^{x} x^{2} - 9 e^{x} x + 9 e^{x}

Add a constant to the solution

= 4 e^{x} x^{2} - 9 e^{x} x + 9 e^{x} + C

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What is the integral of (4x^2-x)e^x ?
The integral of (4x^2-x)e^x is 4e^xx^2-9e^xx+9e^x+C