# what technique do i use to integrate \int(4-x)^{0.3}e^{2x}dx? I have used algebraic manipulation and integration by parts but got no answer. Pls help

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### 1 Answer

You need to use the substitution first, such that:

`2x = t => 2dx = dt => dx = (dt)/2`

`4 - x = 4 - t/2`

Changing the variable yields:

`int (4 - t/2)^(0.3)*e^t*(dt)/2`

You need to use integration by parts such that:

`int udv = uv - int vdu`

`u = (4 - t/2)^(0.3) => du = -(0.3/2)(4 - t/2)^(-0.7)`

`dv = e^t => v= e^t`

`int (4 - t/2)^(0.3)*e^t*(dt)/2 = (1/4)((4 - t/2)^(0.3)*e^t + (0.3/2)int (4 - t/2)^(-0.7)e^t dt)`

Using integration by parts again yields:

`int (4 - t/2)^(-0.7)e^t dt`

`u = e^t => du = e^t dt`

`dv = (4 - t/2)^(-0.7) => v = int u^(-0.7)*(-2du) = -2u^(0.3)/(0.3)`

`int (4 - t/2)^(-0.7)e^t dt = -2e^tu^(0.3)/(0.3) + 2/(0.3)int (4 - t/2)^(0.3)*e^t dt`

You may use the following notation such that:

`int (4 - t/2)^(0.3)*e^t dt = I`

`int (4 - t/2)^(-0.7)e^t dt = -2e^tu^(0.3)/(0.3) + 2/(0.3)*I`

`I= (1/4)((4 - t/2)^(0.3)*e^t + (0.3/2)(-2e^tu^(0.3)/(0.3) + 2/(0.3)*I))`

`I = (1/4)(4 - t/2)^(0.3)*e^t - (e^tu^(0.3))/4 + 1/4I`

You need to isolate to the left side the terms that contain I, such that:

`I - 1/4I =(1/4)(4 - t/2)^(0.3)*e^t - (e^tu^(0.3))/4`

`(3/4)I =(1/4)(4 - t/2)^(0.3)*e^t - (e^tu^(0.3))/4`

`I = (1/3)(4 - t/2)^(0.3)*e^t - (e^tu^(0.3))/3 + c`

**Hence, evaluating the given integral using substitutions and parts, yields `I = (1/3)(4 - t/2)^(0.3)*e^t - (e^tu^(0.3))/3 + c.` **