# Using the method of integration for parts show that ∫xe^(x/2) dx = 2xe^(x/2) - 4e^(x/2) + constant

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### 1 Answer

You need to remember the formula of integration by parts such that:

`int udv = uv - int vdu`

Selecting `u = x ` and `dv = e^(x/2)dx` yields:

`u = x => du = dx`

`dv = e^(x/2)dx => v = (e^(x/2))/(1/2)`

Substituting `dx` for `du ` and `(e^(x/2))/(1/2)` for v yields:

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

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

`int xe^(x/2)dx = 2x(e^(x/2)) - 4(e^(x/2)) + c `

**Hence, evaluating the given integral using parts yields `int xe^(x/2)dx = 2x(e^(x/2)) - 4(e^(x/2)) + c .` **