Calculus of a Single Variable, Chapter 8, 8.2, Section 8.2, Problem 58

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