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

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

Popular >

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

- \frac{1}{2} e^{- 2 x} (1 - x) - (- \frac{1}{4} e^{- 2 x}) : \frac{1}{2} e^{- 2 x} x - \frac{1}{4} e^{- 2 x}

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

Add a constant to the solution

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

\int \frac{cos ( \frac{π}{x ^{21}} )}{x ^{22}} d x

\int x^{2} cos (2 x^{3}) d x

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

\int sin^{4} (\frac{w}{2}) cos^{2} (\frac{w}{2}) d w

\int \frac{ln ( x )}{x ^{m}} d x

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