The integral `int x^n e^-x dx` has to be determined.

Integration by parts gives us the rule: `int u dv = u*v - int v du`

Let `u = x^n` and `dv = e^-x dx`

`du = n*x^(n-1) dx`

`v = -e^(-x)`

`int x^n e^-x dx`

`= x^n*-1*e^-x -...

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The integral `int x^n e^-x dx` has to be determined.

Integration by parts gives us the rule: `int u dv = u*v - int v du`

Let `u = x^n` and `dv = e^-x dx`

`du = n*x^(n-1) dx`

`v = -e^(-x)`

`int x^n e^-x dx`

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

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

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

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

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

Substituting this in the original integral

`int x^n*e^-x dx`

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

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

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

This can be continued n times to yield the final result.

= `-e^-x*(x^n+n*x^(n-1)+ n*(n-1)x^(n-2)+...n!)`

The integral `int x^n*e^(-x)``dx` = `-e^-x*(x^n+n*x^(n-1)+ n*(n-1)x^(n-2)+...n!) + C`

The formula for the integration by parts is as follows:

`int udv = uv - int vdu`

To take the given integral by parts, let

`u = x^n`

and `dv = e^(-x) dx`

Then, `du = nx^(n- 1)` and `v = -e^(-x)`

Plugging these into the formula, we get

`int x^n e^(-x) dx = x^n(-e^(-x)) - int (-e^(-x))nx^(n-1)dx`

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

The resultant integral can be taken by parts again in a similar way, resulting in the following:

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

For a given value of *n*, this process can be repeated *n *times until the power of *x* in the integral is 0. Then, the final integral can be taken as follows:

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

We need to find `intx^n e^(-x)dx`.

Let's solve the problem for `n=1`.

`int xe^(-x)dx=|(u=x, dv=e^(-x)),(du=dx, v=-e^(-x))|=-xe^(-x)+e^(-x)=-e^(-x)(x+1)`` `

Let's now solve the problem for `n=2`.

`int x^2 e^(-x)dx=|(u=x^2, dv=e^(-x)),(du=2xdx, v=-e^(-x))|=-x^2e^(-x)+2int xe^(-x)dx=`

`-x^2e^(-x)-2e^(-x)(x+1)=-e^(-x)(x^2+2x+2)`

Similarly for `n=3` we would have

`int x^3e^(-x)dx=-e^(-x)(x^3+3x^2+6x+6)`

We can now see the pattern

`int x^n e^(-x)dx=-e^(-x) sum_(k=0)^n (x^n)^((k))` **<-- Solution**

where `(x^n)^((k))` is the `k`th derivation of `x^n`. If we were to expand the above formula it would look like this

`int x^n e^(-x)dx=-e^(-x)(x^n+nx^(n-1)+n(n-1)x^(n-2)+cdots+n!x+n!)`

If you wish you can prove the formula formally by using mathematical induction.