Calculus: Early Transcendentals, Chapter 5, 5.5, Section 5.5, Problem 73

You need to solve the definite integral, using fundamental theorem of calculus, such that:
int_a^b f(x) dx = F(b) - F(a)
First, you need to solve the indefinite integral int (dx)/((1+sqrt x)^4) , using the substitution 1 + sqrt x = t such that:
1 + sqrt x = t => 1/(2sqrt x) dx = dt => dx = 2(t-1)dt
int (dx)/((1+sqrt x)^4) = int (2(t-1)dt)/(t^4)
int (2(t-1)dt)/(t^4) = int (2t)/(t^4)dt - int 2/(t^4) dt
int (2(t-1)dt)/(t^4) = int 2/(t^3)dt - int 2/(t^4) dt
int (2(t-1)dt)/(t^4) = int 2*(t^(-3))dt - int 2*(t^(-4)) dt
int (2(t-1)dt)/(t^4) = 2*(t^(-2))/(-2) - 2(t^(-3))/(-3) + c
int (2(t-1)dt)/(t^4) = -1/(t^2) + 2/(3t^3) + c
Replacing back 1 + sqrt x for t yields:
int (dx)/((1+sqrt x)^4) = -1/((1 + sqrt x)^2) + 2/(3(1 + sqrt x)^3) + c
Calculating the integral yields:
int_0^1 (dx)/((1+sqrt x)^4) = (-1/((1 + sqrt x)^2) + 2/(3(1 + sqrt x)^3))|_0^1
int_0^1 (dx)/((1+sqrt x)^4) = (-1/((1 + 1)^2) + 2/(3(1 + 1)^3) + 1/((1 + 0)^2) - 2/(3(1 + 0)^3))
int_0^1 (dx)/((1+sqrt x)^4) = -1/4 + 1/12 + 1 - 2/3
int_0^1 (dx)/((1+sqrt x)^4) = (-3 + 1 + 12 - 8)/12
int_0^1 (dx)/((1+sqrt x)^4) = 2/12
int_0^1 (dx)/((1+sqrt x)^4) = 1/6
Hence, evaluating the definite integral, using the fundamental theorem of calculus, yields int_0^1 (dx)/((1+sqrt x)^4) = 1/6.