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

The integral of (4x+5)e^{-x} is -4e^{-x}x-9e^{-x}+C

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

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

- 4 e^{- x} x - 9 e^{- x} + C

Solution steps

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

Multiply by the conjugate of

4 x + 5 : \frac{16 x ^{2} - 25}{4 x - 5}

= \int \frac{16 x ^{2} - 25}{4 x - 5} e^{- x} d x

Factor

16 x^{2} - 25 : - (- 16 x^{2} + 25)

= \int \frac{- ( - 16 x ^{2} + 25 )}{4 x - 5} e^{- x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{- 16 x ^{2} + 25}{4 x - 5} e^{- x} d x

\frac{- 16 x ^{2} + 25}{4 x - 5} = - 4 x - 5

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

Apply Integration By Parts

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

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

\int 4 e^{- x} d x = - 4 e^{- x}

= - (- e^{- x} (- 4 x - 5) - (- 4 e^{- x}))

Simplify

- (- e^{- x} (- 4 x - 5) - (- 4 e^{- x})) : - 4 e^{- x} x - 9 e^{- x}

= - 4 e^{- x} x - 9 e^{- x}

Add a constant to the solution

= - 4 e^{- x} x - 9 e^{- x} + C

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What is the integral of (4x+5)e^{-x} ?
The integral of (4x+5)e^{-x} is -4e^{-x}x-9e^{-x}+C