# Function/IntegralSolve the indefinite integral of the function (x+1)*(e^-x) .

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may use the property of linearity of integral, hence, you may split the given integral in two simpler integrals, such that:

int (x+1)*(e^-x) dx = int x*e^(-x)dx + int e^(-x) dx

You should use parts to evaluate the integral `int x*e^(-x)dx` such that:

`int f(x)*g'(x)dx = f(x)*g(x) - int f'(x)*g(x) dx`

Considering `f(x) = x` and `g'(x) = e^(-x)` yields:

`{(f(x) = x => f'(x) = 1),(g'(x) = e^(-x)=> g(x) = -e^(-x)):}`

`int x*e^(-x)dx = -x*e^(-x) - int 1*(-e^(-x))dx`

`int x*e^(-x)dx = -x*e^(-x) + int (e^(-x))dx`

`int x*e^(-x)dx = -x*e^(-x) - e^(-x) + c`

`int (x+1)*(e^-x) dx = -x*e^(-x) - e^(-x) - e^(-x) + c`

`int (x+1)*(e^-x) dx = -x*e^(-x) - 2e^(-x) + c`

Hence, evaluating the given indefinite integral yields `int (x+1)*(e^-x) dx = -x*e^(-x) - 2e^(-x) + c.`

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll integrate by parts. For this reason, we'll consider the formula:

Int udv = u*v - Int vdu (*)

We'll put u = x+1. (1)

We'll differentiate both sides:

du = dx (2)

We'll put dv = e^-x (3)

We'll integrate both sides:

Int dv = Int e^-x dx

v = e^-x/-1

v = -e^-x (4)

We'll substitute (1) , (2) , (3) and (4) in (*):

Int udv = -(x+1)*e^-x + Int (e^-x)dx

Int  (x+1)*(e^-x)dx = -(x+1)*e^-x  - e^-x + C

Int  (x+1)*(e^-x)dx = -(e^-x)*(x+2) + C