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

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

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

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

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

Solution steps

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

Factor

1 - e^{t} : - (- 1 + e^{t})

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

(- (- 1 + e^{t}))^{2} = (- 1 + e^{t})^{2}

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

Apply u-substitution

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

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

Expand

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

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

\int 1 d u = u

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

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

= u - u^{2} + \frac{u ^{3}}{3}

Substitute back

u = e^{t}

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

Simplify

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

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

Add a constant to the solution

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

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