Evaluate the integral of 1/(1+sqrtx)^4 from [0,1]

Question

I started off letting u=1+sqrtx and du=1/2sqrtx, but then there is no 1/2sqrtx in the equation. What am I suppose to do next?

The answer in the book is 1/6.

Answer 1

u = 1+√x
√x = u-1
du = 1/(2√x) dx = 1/(2u-2) dx
so, dx = (2u-2) du

∫1/(1+√x)^4 dx
= ∫1/u^4 (2u-2) du
= 2∫(1/u^3 - 1/u^4) du
...