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

| Certified Educator

You need to use the linearity of integral such that:

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

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

You need to solve the integral `int 1/x*e^(-x)dx` using parts such that:

`int udv = uv - int vdu`

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

`dv = e^(-x)dx => 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^2 e^x) dx = e^(-x) + 2(e^(-x))/x + 2int 1/x^2*e^(-x)dx- 2 int 1/(x^2)*e^(-x)dx`

Reducing like terms yields:

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

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

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

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