What is the integral of x^2(e^x-1) ?

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

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integral of x^2(e^x-1)

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

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

Solution steps

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

Expand

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

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

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

\int x^{2} d x = \frac{x ^{3}}{3}

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

Add a constant to the solution

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

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