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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 -

    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

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