For the region bounded by `y=3/(1+x)` , `y=0` , and `x=3` revolved
about the line `x=4` , we may apply **Washer method** for
the integral application for the volume of a solid. Note: We consider a region
bounded at the first quadrant.

The formula for the Washer Method is:

`V...

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= pi int_a^b [(f(x))^2-(g(x))^2]dx`

or

`V = pi int_a^b [(f(y))^2-(g(y))^2]dy`

where *f as function of the outer radius *

*g as a function of the inner
radius*

To determine which form we use, we consider the vertical rectangular strip representation that is perpendicular to the axis of rotation as shown on the attached image. The given strip will have a thickness of "`dx` " which is our clue to use the formula:

`V = pi int_a^b [(f(x))^2-(g(x))^2]dx`

For each radius, we follow the `y_(above)-y_(below)` .

For the **inner radius**, we have: `g(x) = 4-3/(1+x)`

For the **outer radius**, we have: `f(x) = 4-0 =4`

Then the **boundary values of x** is ` a=0` and `b
=3` .

The integral will be:

`V = pi int_0^3 [4^2 -(4-3/(1+x))^2]dx`

Expand using the FOIL method on:`(4-3/(1+x))^2= 9/(x + 1)^2 - 24/(x + 1) + 16` and `4^2 =16` .

The integral becomes:

`V = pi int_0^3 [16 -(9/(x + 1)^2 - 24/(x + 1) + 16)]dx`

`V = pi int_0^3 [16 -9/(x + 1)^2 + 24/(x + 1) -16]dx`

Simplify: `V = pi int_0^3 [-9/(x + 1)^2+ 24/(x + 1) ]dx` or

`V = pi int_0^3 [ 24/(x + 1) -9/(x + 1)^2 ]dx`

Apply basic integration property:

`int (u-v)dx = int (u)dx-int (v)dx.`

`V = pi [int_0^3 24/(x + 1)dx -int_0^3 9/(x + 1)^2dx ]`

For the indefinite integral, may apply u-substitution by letting: `u=x+1` then `du=dx`

`V = pi [int 24/(u)du -int 9/(u)^2du]`

For the integral of `int 24/(u)du` , we follow the integration formula for logarithm:

`int 24/(u)du= 24ln|u|`

For the integral of `int 9/(u)^2du` , we follow the Law of Exponent: `1/x^n = x^-n` and Power rule for integration formula : `int x^n dx =x^(n+1)/(n+1)` .

`int 9/(u)^2du =int 9u^(-2)du`

` = 9u^(-2+1)/(-2+1)`

` = 9 u^(-1)/(-1)`

`= -9/u`

Then, the integral becomes:

`V = pi [int 24/(u)du -int 9/(u)^2du]`

`V = pi [ 24ln|u| -(-9/u)]`

`V = pi [ 24ln|u| +9/u]`

Plug-in `u=x+1` on `V = pi [ 24ln|u| +9/u]` , we get:

`V = pi [ 24ln|(x+1)| +9/(x+1)]|_0^3`

Apply the definite integral formula: `int _a^b f(x) dx = F(b) - F(a)` .

`V = pi [ 24ln|(3+1)| +9/(3+1)] -pi [ 24ln|(0+1)| +9/(0+1)]`

`V = pi [ 24ln|4| +9/4] -pi [ 24ln|(1)| +9/(1)]`

`V = pi [ 24ln|4| +9/4] -pi [ 24*0 +9]`

`V = pi [ 24ln|4| +9/4 -9]`

`V = pi [ 24ln|4| -27/4 ]`