What is the integral of (2x+6)e^{-4x} ?

The integral of (2x+6)e^{-4x} is -1/2 e^{-4x}x-13/8 e^{-4x}+C

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integral of (2x+6)e^{-4x}

\int (2 x + 6) e^{- 4 x} d x

- \frac{1}{2} e^{- 4 x} x - \frac{13}{8} e^{- 4 x} + C

Solution steps

\int (2 x + 6) e^{- 4 x} d x

Apply Integration By Parts

: - \frac{1}{2} e^{- 4 x} (x + 3) - \int - \frac{1}{2} e^{- 4 x} d x

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

\int - \frac{1}{2} e^{- 4 x} d x = \frac{1}{8} e^{- 4 x}

= - \frac{1}{2} e^{- 4 x} (x + 3) - \frac{1}{8} e^{- 4 x}

Simplify

- \frac{1}{2} e^{- 4 x} (x + 3) - \frac{1}{8} e^{- 4 x} : - \frac{1}{2} e^{- 4 x} x - \frac{13}{8} e^{- 4 x}

= - \frac{1}{2} e^{- 4 x} x - \frac{13}{8} e^{- 4 x}

Add a constant to the solution

= - \frac{1}{2} e^{- 4 x} x - \frac{13}{8} e^{- 4 x} + C

\frac{\partial}{\partial x} (cos (5 x))

f (x) = (8 x + 9) (9 x - 10)

\frac{d}{d x} (7 (cos (x^{2}) - 2 x^{2} sin (x^{2})))

f (x) = x^{2} 2^{x}

\int \frac{5}{x ^{2} + 4 x - 1} d x

What is the integral of (2x+6)e^{-4x} ?
The integral of (2x+6)e^{-4x} is -1/2 e^{-4x}x-13/8 e^{-4x}+C