What is the integral of (1-6x)e^{2x} ?

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

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integral of (1-6x)e^{2x}

\int (1 - 6 x) e^{2 x} d x

- 3 e^{2 x} x + 2 e^{2 x} + C

Solution steps

\int (1 - 6 x) e^{2 x} d x

Apply Integration By Parts

: \frac{1}{2} e^{2 x} (1 - 6 x) - \int - 3 e^{2 x} d x

= \frac{1}{2} e^{2 x} (1 - 6 x) - \int - 3 e^{2 x} d x

\int - 3 e^{2 x} d x = - \frac{3}{2} e^{2 x}

= \frac{1}{2} e^{2 x} (1 - 6 x) - (- \frac{3}{2} e^{2 x})

Simplify

\frac{1}{2} e^{2 x} (1 - 6 x) - (- \frac{3}{2} e^{2 x}) : - 3 e^{2 x} x + 2 e^{2 x}

= - 3 e^{2 x} x + 2 e^{2 x}

Add a constant to the solution

= - 3 e^{2 x} x + 2 e^{2 x} + C

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