What is the integral of (x^2-2x+5)e^x ?

The integral of (x^2-2x+5)e^x is e^xx^2-4e^xx+9e^x+C

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integral of (x^2-2x+5)e^x

\int (x^{2} - 2 x + 5) e^{x} d x

e^{x} x^{2} - 4 e^{x} x + 9 e^{x} + C

Solution steps

\int (x^{2} - 2 x + 5) e^{x} d x

Expand

(x^{2} - 2 x + 5) e^{x} : e^{x} x^{2} - 2 e^{x} x + 5 e^{x}

= \int e^{x} x^{2} - 2 e^{x} x + 5 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 2 e^{x} x d x + \int 5 e^{x} d x

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

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

\int 5 e^{x} d x = 5 e^{x}

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

Simplify

x^{2} e^{x} - 2 (e^{x} x - e^{x}) - 2 (e^{x} x - e^{x}) + 5 e^{x} : e^{x} x^{2} - 4 e^{x} x + 9 e^{x}

= e^{x} x^{2} - 4 e^{x} x + 9 e^{x}

Add a constant to the solution

= e^{x} x^{2} - 4 e^{x} x + 9 e^{x} + C

\int cos (- 2 x) d x

\frac{d x}{d t} + x = 3 e^{t}

y^{^{'}} - y = e^{3 x}

f (x) = 11 x^{4} cot (x)

\frac{d}{d x} (x (5 x + 1)^{\frac{1}{3}})

What is the integral of (x^2-2x+5)e^x ?
The integral of (x^2-2x+5)e^x is e^xx^2-4e^xx+9e^x+C