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

The integral of (e^{3x}+2)^2 is 1/3 (1/2 e^{6x}+4e^{3x}+12x)+C

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

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

\frac{1}{3} (\frac{1}{2} e^{6 x} + 4 e^{3 x} + 12 x) + C

Solution steps

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

Apply u-substitution

: \int \frac{( e ^{u} + 2 ) ^{2}}{3} d u

= \int \frac{( e ^{u} + 2 ) ^{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 (e^{u} + 2)^{2} d u

Expand

(e^{u} + 2)^{2} : e^{2 u} + 4 e^{u} + 4

= \frac{1}{3} \cdot \int e^{2 u} + 4 e^{u} + 4 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

= \frac{1}{3} (\int e^{2 u} d u + \int 4 e^{u} d u + \int 4 d u)

\int e^{2 u} d u = \frac{1}{2} e^{2 u}

\int 4 e^{u} d u = 4 e^{u}

\int 4 d u = 4 u

= \frac{1}{3} (\frac{1}{2} e^{2 u} + 4 e^{u} + 4 u)

Substitute back

u = 3 x

= \frac{1}{3} (\frac{1}{2} e^{2 \cdot 3 x} + 4 e^{3 x} + 4 \cdot 3 x)

Simplify

= \frac{1}{3} (\frac{1}{2} e^{6 x} + 4 e^{3 x} + 12 x)

Add a constant to the solution

= \frac{1}{3} (\frac{1}{2} e^{6 x} + 4 e^{3 x} + 12 x) + C

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What is the integral of (e^{3x}+2)^2 ?
The integral of (e^{3x}+2)^2 is 1/3 (1/2 e^{6x}+4e^{3x}+12x)+C