What is the integral of (x^2+x+1)(e^x) ?

The integral of (x^2+x+1)(e^x) is e^xx^2-e^xx+2e^x+C

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integral of (x^2+x+1)(e^x)

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

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

Solution steps

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

Apply Integration By Parts

= (x^{2} + x + 1) e^{x} - \int (2 x + 1) e^{x} d x

\int (2 x + 1) e^{x} d x = 2 e^{x} x - e^{x}

= (x^{2} + x + 1) e^{x} - (2 e^{x} x - e^{x})

Simplify

(x^{2} + x + 1) e^{x} - (2 e^{x} x - e^{x}) : e^{x} x^{2} - e^{x} x + 2 e^{x}

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

Add a constant to the solution

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

What is the integral of (x^2+x+1)(e^x) ?
The integral of (x^2+x+1)(e^x) is e^xx^2-e^xx+2e^x+C