∫dx/(2 √x +2x)

I tried∫1/(2 √x +2x)dx

then i know that it is not du/u so i tried finding u' but i don't think i have the right one...

i found 6x^1/2 because the bottom is 4x^3/2

so: -6x^1/2∫6x^1/2/(2 √x +2x)dx

then: -6x^1/2[ln(abs value 4x^3/2)+c]

but i don't know after that.

the answer is ln(1+ √x) +c