# what is the integration of (e^x)/(x^3) dx?

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### 1 Answer

Integrate by parts using `dv = 1/x^3 dx` and `u = e^x` we get

`v = -1/(2x^2)` and `du = e^x dx`

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

Now `int e^x (-1/(2x^2)) dx = -1/2 int e^x/x^2 dx` and we use integration by parts again to get

`dv = 1/x^2 dx` and `u = e^x` so `v = -1/x` and `u = e^x` still and we get

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

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

Since `int e^(-x)/x dx = -Ei(x)` Exponential integral we get

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

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