∫ 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