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

The integral of e^{3x}(1-e^{3x})^2 is -1/9 (1-e^{3x})^3+C

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integral of e^{3x}(1-e^{3x})^2

\int e^{3 x} (1 - e^{3 x})^{2} d x

- \frac{1}{9} (1 - e^{3 x})^{3} + C

Solution steps

\int e^{3 x} (1 - e^{3 x})^{2} d x

Apply u-substitution

: \int - \frac{u ^{2}}{3} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{3} \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= - \frac{1}{3} \cdot \frac{u ^{3}}{3}

Substitute back

u = 1 - e^{3 x}

= - \frac{1}{3} \cdot \frac{( 1 - e ^{3 x} ) ^{3}}{3}

Simplify

- \frac{1}{3} \cdot \frac{( 1 - e ^{3 x} ) ^{3}}{3} : - \frac{1}{9} (1 - e^{3 x})^{3}

= - \frac{1}{9} (1 - e^{3 x})^{3}

Add a constant to the solution

= - \frac{1}{9} (1 - e^{3 x})^{3} + C

\frac{d}{d x} (\frac{e ^{x}}{x ^{2} + 1})

tan (3 x), 2 sin (3 x), [- \frac{π}{9}, \frac{π}{9}]

\frac{\partial}{\partial y} (e^{x^{2} y})

y = \frac{x - 1}{x + 1}

\int \frac{( 4 x - 11 )}{( x ^{3} - 9 x ^{2} )} d x

What is the integral of e^{3x}(1-e^{3x})^2 ?
The integral of e^{3x}(1-e^{3x})^2 is -1/9 (1-e^{3x})^3+C