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

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`