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

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

Popular >

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

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

e (- e^{- x} x - 2 e^{- x}) + C

Solution steps

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

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

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

Take the constant out:

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

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

Simplify

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

Apply Integration By Parts

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

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

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

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

Expand

- e^{- x} (x + 1) - e^{- x} : - e^{- x} x - 2 e^{- x}

= e (- e^{- x} x - 2 e^{- x})

Add a constant to the solution

= e (- e^{- x} x - 2 e^{- x}) + C

\int \frac{ln ( 6 x )}{x ^{7}} d x

\int (x^{2} + 2 x + 1)^{10} d x

\int (6 x^{2} + 5 x + \frac{1}{x ^{3}}) d x

\int 1 0^{4 x - 2} d x

\int \frac{3 x + 1}{( 3 x ^{2} + 2 x + 1 ) ^{6}} d x

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