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

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

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

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

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

Solution steps

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

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