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

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

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integral of x(x+1)e^x

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

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

Solution steps

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

Apply Integration By Parts

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

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

Simplify

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

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

Add a constant to the solution

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

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