What is the integral of (x+6)e^{3x} ?

The integral of (x+6)e^{3x} is 1/3 e^{3x}x+17/9 e^{3x}+C

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integral of (x+6)e^{3x}

\int (x + 6) e^{3 x} d x

\frac{1}{3} e^{3 x} x + \frac{17}{9} e^{3 x} + C

Solution steps

\int (x + 6) e^{3 x} d x

Apply Integration By Parts

: \frac{1}{3} e^{3 x} (x + 6) - \int \frac{1}{3} e^{3 x} d x

= \frac{1}{3} e^{3 x} (x + 6) - \int \frac{1}{3} e^{3 x} d x

\int \frac{1}{3} e^{3 x} d x = \frac{1}{9} e^{3 x}

= \frac{1}{3} e^{3 x} (x + 6) - \frac{1}{9} e^{3 x}

Simplify

\frac{1}{3} e^{3 x} (x + 6) - \frac{1}{9} e^{3 x} : \frac{1}{3} e^{3 x} x + \frac{17}{9} e^{3 x}

= \frac{1}{3} e^{3 x} x + \frac{17}{9} e^{3 x}

Add a constant to the solution

= \frac{1}{3} e^{3 x} x + \frac{17}{9} e^{3 x} + C

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What is the integral of (x+6)e^{3x} ?
The integral of (x+6)e^{3x} is 1/3 e^{3x}x+17/9 e^{3x}+C