how to integrate    ( x^2+2x+2)/(x^3 e^x)  

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You need to use the linearity of integrals such that:

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

`int (x^2+2x+2)/(x^3 e^x) dx = int 1/x e^(-x)dx + 2 int 1/x^2 e^(-x)dx + 2 int 1/x^3 e^(-x)dx`

You need to use integration by parts to solve `int 1/x e^(-x)dx` such that:

`int udv = uv - int vdu`

`u = 1/x => du = -1/x^2`

`dv = e^(-x) => v = -e^(-x)`

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

`int (x^2+2x+2)/(x^3 e^x) dx = -e^(-x)/x - int 1/x^2 e^(-x) dx + 2 int 1/x^2 e^(-x)dx + 2 int 1/x^3 e^(-x)dx`

`int (x^2+2x+2)/(x^3 e^x) dx = -e^(-x)/x + int 1/x^2 e^(-x)dx + 2 int 1/x^3 e^(-x)dx `

You need to use integration by parts to solve `int 1/x^3 e^(-x)dx`  such that:

`int 1/x^3 e^(-x)dx = -1/(2x^2) e^(-x) - (1/2)int 1/x^2 e^(-x)dx `

`u = e^(-x) => du = -e^(-x)`

`dv = 1/x^3 => v = -1/(2x^2)`

`int (x^2+2x+2)/(x^3 e^x) dx = -e^(-x)/x + int 1/x^2 e^(-x)dx -1/(x^2) e^(-x) - int 1/x^2 e^(-x)dx`

Reducing like terms yields:

`int (x^2+2x+2)/(x^3 e^x) dx = -e^(-x)/x -1/(x^2) e^(-x) + c`

`int (x^2+2x+2)/(x^3 e^x) dx = -e^(-x)/x(1 + 1/x) + c`

Hence, evaluating the given integral, using mostly integration by parts, `int (x^2+2x+2)/(x^3 e^x) dx = -e^(-x)/x(1 + 1/x) + c.`

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