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

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

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

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

e^{x} x^{2} - 3 (e^{x} x - e^{x}) + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

y^{^{'}} = y (1 - y), y (0) = 12

\int \frac{1}{x ^{2} 3 x - 1} d x

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

\frac{\partial}{\partial y} (y sin (x y))

\frac{d}{d x} ((ln (x))^{5})

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