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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int u - 1 d u

= \int u - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u d u - \int 1 d u

\int u d u = \frac{u ^{2}}{2}

\int 1 d u = u

= \frac{u ^{2}}{2} - u

Substitute back

u = e^{x}

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

Simplify

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

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

Add a constant to the solution

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

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