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

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

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

\int \frac{x ^{2} + x}{e ^{x}} d x

- \frac{x ^{2}}{e ^{x}} - \frac{2 x}{e ^{x}} - \frac{x}{e ^{x}} - \frac{3}{e ^{x}} + C

Solution steps

\int \frac{x ^{2} + x}{e ^{x}} d x

Expand

\frac{x ^{2} + x}{e ^{x}} : \frac{x ^{2}}{e ^{x}} + \frac{x}{e ^{x}}

= \int \frac{x ^{2}}{e ^{x}} + \frac{x}{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 \frac{x ^{2}}{e ^{x}} d x + \int \frac{x}{e ^{x}} d x

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

\int \frac{x}{e ^{x}} d x = - \frac{x}{e ^{x}} - \frac{1}{e ^{x}}

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

Simplify

- \frac{x ^{2}}{e ^{x}} + 2 (- \frac{x}{e ^{x}} - \frac{1}{e ^{x}}) - \frac{x}{e ^{x}} - \frac{1}{e ^{x}} : - \frac{x ^{2}}{e ^{x}} - \frac{2 x}{e ^{x}} - \frac{x}{e ^{x}} - \frac{3}{e ^{x}}

= - \frac{x ^{2}}{e ^{x}} - \frac{2 x}{e ^{x}} - \frac{x}{e ^{x}} - \frac{3}{e ^{x}}

Add a constant to the solution

= - \frac{x ^{2}}{e ^{x}} - \frac{2 x}{e ^{x}} - \frac{x}{e ^{x}} - \frac{3}{e ^{x}} + C

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