Let x = t 2 and dx=2tdt.
∫[ 1/( 1+√x)]dx =∫[2t/( 1+t)]dt
= 2∫[t/( 1+t)]dt = 2∫[( 1+t- 1)/( 1+t)]dt
= 2∫{ 1-[ 1/( 1+t)]} dt = 2t-2ln | 1+t |+C
And because t=√x, the original formula = 2 √ x-2ln |1+√ x |+c.
Because the original problem is to calculate the definite integral, the upper and lower limits of the two integrals of 4 and 0 are substituted for calculation:
Original integral =2√4-2ln| 1+√4|=4-2ln3