What is the integral of (3-2x)e^{5-4x} ?

The integral of (3-2x)e^{5-4x} is 1/2 e^{-4x+5}x-5/8 e^{-4x+5}+C

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integral of (3-2x)e^{5-4x}

\int (3 - 2 x) e^{5 - 4 x} d x

\frac{1}{2} e^{- 4 x + 5} x - \frac{5}{8} e^{- 4 x + 5} + C

Solution steps

\int (3 - 2 x) e^{5 - 4 x} d x

Expand

(3 - 2 x) e^{5 - 4 x} : 3 e^{5 - 4 x} - 2 e^{5 - 4 x} x

= \int 3 e^{5 - 4 x} - 2 e^{5 - 4 x} x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 3 e^{5 - 4 x} d x - \int 2 e^{5 - 4 x} x d x

\int 3 e^{5 - 4 x} d x = - \frac{3}{4} e^{5 - 4 x}

\int 2 e^{5 - 4 x} x d x = 2 (- \frac{1}{4} e^{5 - 4 x} x - \frac{1}{16} e^{5 - 4 x})

= - \frac{3}{4} e^{5 - 4 x} - 2 (- \frac{1}{4} e^{5 - 4 x} x - \frac{1}{16} e^{5 - 4 x})

Simplify

- \frac{3}{4} e^{5 - 4 x} - 2 (- \frac{1}{4} e^{5 - 4 x} x - \frac{1}{16} e^{5 - 4 x}) : \frac{1}{2} e^{- 4 x + 5} x - \frac{5}{8} e^{- 4 x + 5}

= \frac{1}{2} e^{- 4 x + 5} x - \frac{5}{8} e^{- 4 x + 5}

Add a constant to the solution

= \frac{1}{2} e^{- 4 x + 5} x - \frac{5}{8} e^{- 4 x + 5} + C

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What is the integral of (3-2x)e^{5-4x} ?
The integral of (3-2x)e^{5-4x} is 1/2 e^{-4x+5}x-5/8 e^{-4x+5}+C