Verified answer

∫dx/(1 + √(2x))

u = √(2x) + 1

(u - 1)²/2 = x

(u - 1) du = dx

∫(u - 1)/u du

∫(1 - 1/u) du = u - ln|u| + C

√(2x) - ln|√(2x) + 1| + C

Note that the +1 gets absorbed into the +C.

Let u = √2x. Then u² = 2x and 2u du = 2 dx. Then u du = dx and

∫ 1/(1 + √(2x)) dx = ∫u/(1 + u) du = ∫(1 - 1/(1 + u)) du = u - ln|1 + u| + C

= √2x - ln|1 + √2x| + C.