Math
posted by Dave .
Evaluate the integral of 1/(1+sqrtx)^4 from [0,1]
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.

Math 
Steve
u = 1+√x
√x = u1
du = 1/(2√x) dx = 1/(2u2) dx
so, dx = (2u2) du
∫1/(1+√x)^4 dx
= ∫1/u^4 (2u2) du
= 2∫(1/u^3  1/u^4) du
...
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