What is the integral of (9x-1)e^{-x} ?

The integral of (9x-1)e^{-x} is -9e^{-x}x-8e^{-x}+C

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integral of (9x-1)e^{-x}

\int (9 x - 1) e^{- x} d x

- 9 e^{- x} x - 8 e^{- x} + C

Solution steps

\int (9 x - 1) e^{- x} d x

Apply Integration By Parts

: - e^{- x} (9 x - 1) - \int - 9 e^{- x} d x

= - e^{- x} (9 x - 1) - \int - 9 e^{- x} d x

\int - 9 e^{- x} d x = 9 e^{- x}

= - e^{- x} (9 x - 1) - 9 e^{- x}

Simplify

- e^{- x} (9 x - 1) - 9 e^{- x} : - 9 e^{- x} x - 8 e^{- x}

= - 9 e^{- x} x - 8 e^{- x}

Add a constant to the solution

= - 9 e^{- x} x - 8 e^{- x} + C

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

\int sin (x) \cdot sin (2 \cdot x) d x

\frac{x - x}{x ^{\frac{1}{4}}}

\frac{5}{2} x^{\frac{5}{2}}

f (x) = 2 x^{4} - 10 x^{2}, at x = - 2

What is the integral of (9x-1)e^{-x} ?
The integral of (9x-1)e^{-x} is -9e^{-x}x-8e^{-x}+C