integrate sqrt(2x)
1) Let's integrate: ∫(2x)dx
For 2x, we use integration by parts: ∫(u1·v2')dx = u1·v2 + ∫(u1'·v2)dx, with u1 = 2x and v2' = 1
2) Let's perform the differentiation u1: (2x)'
We differentiate using: (f)' = (f)'·12f (where f = 2x)
(2x)'·12·2x
22·2x
12x
Thus, u1' is 12x
3) Let's compute the integral v2': ∫(1)dx
x
4) Let's evaluate u1'·v2: 12x·x
x2x
x2·x
We apply simplification rules, we can replace the fraction: xx with the new expression: x(1 -12)
x(1 -12)2
x(2 -12)2
x2
The conventional preference is to have numerical square roots in the numerator rather than in the denominator:
x·22
5) Let's compute the integral u1·v2 + ∫(u1'·v2)dx: 2x·x -∫(x·22)dx
Use the integration formula for x·22: ∫(xn)dx = x(n + 1)n + 1
2x·x -22·x(12 + 1)12 + 1
2x·x -2·x(1 + 22)2·1 + 22
2x·x -2·x322·32
2x·x -2·x3262
2x·x -2·x32·13
2x·x -2·x323
Indefinite integrals are defined up to an additive constant C, so this yields: 2x·x -2·x323 + C
Answer:   -x32·23 + x·2x + C, with C being any constant