# use integration by parts to find the indefinate integral Int_ (x-1)/(e^x) dx

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### 2 Answers

You should rewrite the integrand such that:

`(x - 1)/(e^x) = (x - 1)*e^(-x)`

You need to integrate using parts, hence you need to use the following formula such that:

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

Select the function `f(x) = x-1` , hence, differentiating f(x), the degree of polynomial decreases such that:

`f(x) = x - 1 => (df(x))/dx = 1`

`g'(x) = e^(-x) dx => g(x) = -e^(-x)`

Substituting `f(x), f'(x), g(x) , g'(x)` into the formula above yields:

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

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

**Hence, integrating using parts yields `int (x - 1)*e^(-x) dx = - x*e^(-x) + c.` **

**Sources:**

The integration by parts formula is `int udv=uv-int vdu`

Let `u=x` and `dv=e^{-x}dx` . Then `du=dx` and `v=-e^{-x}` .

Then the integral becomes

`int{x-1}/e^xdx`

`=int x/e^x dx-int 1/e^xdx`

`=int xe^{-x}dx-int e^{-x} dx` apply integ by parts to first integral

`=-xe^{-x} +int e^{-x}dx - int e^{-x}dx + C` where C is constant

`=-xe^{-x}+C`

**The integral is `-xe^{-x}+C` .**