# 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 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

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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.

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