`int_0^1 (dx)/(1 + sqrt(x))^4` Evaluate the definite integral.

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)...

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.`