For the region bounded by `y=2/(x+1)` ,`y=0` , `x=0` and `x=6` and revolved about the x-axis, we may apply **Disk method**. For the Disk method, we consider a** perpendicular rectangular strip with the axis of revolution**.

As shown on the attached image, the* thickness of the rectangular strip is "dx"* with a vertical orientation perpendicular to the x-axis (axis of revolution).

We follow the formula for the Disk method:`V = int_a^b A(x) dx` where disk base area is `A= pi r^2` with` r =y=f(x)` .

For the r, we consider the *length of the strip*`= y_(above) - y_(below).`

Then `r =f(x)= 2/(x+1)-0`

` f(x) = 2/(x+1)`

The boundary values of x will be `a=0` to `b=6` .

Plug-in the `f(x)` and the boundary values to integral formula, we get:

`V = int_0^6 pi (2/(x+1))^2 dx`

`V = int_0^6 (4pi)/(x+1)^2 dx`

To solve for the indefinite integral, we may apply u-substitution by using` u = x+1` then `du =dx` .

The integral becomes:

`V = int (4pi)/(u^2) du`

Apply basic integration property: `int c f(x) dx = c int f(x) dx` .

`V =4pi int 1/u^2 du`

Apply Law of exponent: `1/x^n = x^(-n)` and Power rule of integration:` int x^n dx = x^(n+1)/(n+1)`

`V =4pi int u^(-2) du`

`V= 4pi u^((-2+1))/((-2+1))`

`V=4pi* u^(-1)/(-1)`

`V=(-4pi)/u`

Plug-in `u=x+1` on` V=(-4pi)/u` , we get:

`V=(-4pi)/(x+1)` with boundary values `a=0` to `b =6` .

Apply definite integration formula: `int_a^b f(y) dy= F(b)-F(a)` .

`V = (-4pi)/(6+1) -(-4pi)/(0+1)`

`V =(-4pi)/7 +4pi)`

`V =(24pi)/7` or `10.77` (approximated value)