What is the integral of (3x^2+7x+1)e^x ?

The integral of (3x^2+7x+1)e^x is 3e^xx^2+e^xx+C

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integral of (3x^2+7x+1)e^x

\int (3 x^{2} + 7 x + 1) e^{x} d x

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

Solution steps

\int (3 x^{2} + 7 x + 1) e^{x} d x

Apply Integration By Parts

= (3 x^{2} + 7 x + 1) e^{x} - \int (6 x + 7) e^{x} d x

\int (6 x + 7) e^{x} d x = 6 e^{x} x + e^{x}

= (3 x^{2} + 7 x + 1) e^{x} - (6 e^{x} x + e^{x})

Simplify

(3 x^{2} + 7 x + 1) e^{x} - (6 e^{x} x + e^{x}) : 3 e^{x} x^{2} + e^{x} x

= 3 e^{x} x^{2} + e^{x} x

Add a constant to the solution

= 3 e^{x} x^{2} + e^{x} x + C

What is the integral of (3x^2+7x+1)e^x ?
The integral of (3x^2+7x+1)e^x is 3e^xx^2+e^xx+C