`(dy)/(dx)=(-8)/((2x+3)^3)`

`dy=(-8)/((2x+3)^3)dx` Integrate both sides:

`y=int (-8)/((2x+3)^3)dx`

Let `u=2x+3` ; then `du=2dx` Then we have:

`y=-4 int (du)/u^3=-4 int u^(-3)du`

`y=-4[-1/2 u^(-2)+C_1]` ** `C_1` a constant

`y=2u^(-2)+C` ** `C=-4C_1` ** Substitute for u:

`y=2(2x+3)^(-2)+C` or `y=2/((2x+3)^2)+C`

Now we have the point (-1,3), so we can solve for the constant:

`3=2/((-2+3)^2)+C`

`3=2+C=>C=1`

-----------------------------------------

**Thus the solution is `y=2/((2x+3)^2)+1` **

-----------------------------------------

To check, we can differentiate:

`(dy)/(dx)=d/(dx)[2(2x+3)^(-2)+1]`

`=-4(2x+3)^(-3)(2)` Using the product and chain rules

`=-8(2x+3)^(-3)`

`=(-8)/(2x+3)^(3)` as required.

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.