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

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

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integral of (1-e^x)^2

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

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

Solution steps

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

Apply u-substitution

: \int \frac{( 1 - u ) ^{2}}{u} d u

= \int \frac{( 1 - u ) ^{2}}{u} d u

Expand

\frac{( 1 - u ) ^{2}}{u} : \frac{1}{u} - 2 + u

= \int \frac{1}{u} - 2 + u 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 \frac{1}{u} d u - \int 2 d u + \int u d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int 2 d u = 2 u

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

= ln ∣ u ∣ - 2 u + \frac{u ^{2}}{2}

Substitute back

u = e^{x}

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

Simplify

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

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

Add a constant to the solution

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

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