What is the integral of e^x(x^2+3x-4) ?

The integral of e^x(x^2+3x-4) is e^xx^2+e^xx-5e^x+C

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integral of e^x(x^2+3x-4)

\int e^{x} (x^{2} + 3 x - 4) d x

e^{x} x^{2} + e^{x} x - 5 e^{x} + C

Solution steps

\int e^{x} (x^{2} + 3 x - 4) d x

Expand

e^{x} (x^{2} + 3 x - 4) : e^{x} x^{2} + 3 e^{x} x - 4 e^{x}

= \int e^{x} x^{2} + 3 e^{x} x - 4 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^{x} x^{2} d x + \int 3 e^{x} x d x - \int 4 e^{x} d x

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

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

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

= x^{2} e^{x} - 2 (e^{x} x - e^{x}) + 3 (e^{x} x - e^{x}) - 4 e^{x}

Simplify

= e^{x} x^{2} + e^{x} x - 5 e^{x}

Add a constant to the solution

= e^{x} x^{2} + e^{x} x - 5 e^{x} + C

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What is the integral of e^x(x^2+3x-4) ?
The integral of e^x(x^2+3x-4) is e^xx^2+e^xx-5e^x+C