What is the integral of (e^x-1)^3 ?

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

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integral of (e^x-1)^3

\int (e^{x} - 1)^{3} d x

\frac{1}{3} e^{3 x} - \frac{3}{2} e^{2 x} + 3 e^{x} - x + C

Solution steps

\int (e^{x} - 1)^{3} d x

Expand

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

= \int e^{3 x} - 3 e^{2 x} + 3 e^{x} - 1 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^{3 x} d x - \int 3 e^{2 x} d x + \int 3 e^{x} d x - \int 1 d x

\int e^{3 x} d x = \frac{1}{3} e^{3 x}

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

\int 3 e^{x} d x = 3 e^{x}

\int 1 d x = x

= \frac{1}{3} e^{3 x} - \frac{3}{2} e^{2 x} + 3 e^{x} - x

Add a constant to the solution

= \frac{1}{3} e^{3 x} - \frac{3}{2} e^{2 x} + 3 e^{x} - x + C

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