`int e^(sqrt(2x)) dx` Find the indefinite integral by using substitution followed by integration by parts.

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To evaluate the given integral problem` int e^(sqrt(2x))dx ` us u-substituion, we may let:

`u = 2x` then `du = 2 dx` or `(du)/2 = dx` .

Plug-in the values `u = 2x ` and `dx = (du)/2` , we get:

`int e^(sqrt(2x))dx =int e^(sqrt(u))* (du)/2`

Apply the basic integration property: `int c*f(x) dx = c int f(x) dx` .

`int e^(sqrt(u))* (du)/2=1/2 int e^(sqrt(u)) du`

Apply another set of substitution, we let:

`w = sqrt(u)`

Square both sides of `w =sqrt(u)`, we get: `w^2 =u`

Take the derivative on each side, it becomes: `2w dw = du` 

Plug-in `w =sqrt(u)` and `du = 2w dw` , we get: 

`1/2 int e^(sqrt(u)) du =1/2 int e^(w) * 2w dw`

                                     ` = 1/2 * 2 inte^(w) *w dw`

                                     `= int e^w * w dw` .

To evaluate the integral further, we apply integration by parts:`int f* g' = f*g - int g *f'

Let: `f =w` then `f' = dw`

       `g' = e^w dw` then `g = e^w`

Applying the formula for integration by parts, we get:

`int e^w * w dw = w*e^w - int e^w dw`

                       `= we^w -e^w +C`

Recall we let: `w =sqrt(u)` and `u = 2x ` then `w =sqrt(2x)` .

 Plug-in `w=sqrt(2x)` on  `we^w -e^w +C` , we get the complete indefinite integral as:

`int e^(sqrt(2x))dx =sqrt(2x) e^(sqrt(2x)) -e^(sqrt(2x)) +C`

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