To solve the differential equation of `(dy)/(dx)=(x^3-2x)/(5+4x-x^2)` .we may express it in a form of **variable separable differential equation**:`N(y) dy = M(x) dx`

`dy=(x^3-2x)/(5+4x-x^2) dx`

Then apply **direct integration** on both sides:

`int dy= int (x^3-2x)/(5+4x-x^2) dx`

For the left side, we apply the basis integration property:` int dy = y`

For the right side, we may apply long division to expand:

`int (x^3-2x)/(5+4x-x^2)dx= int [-x-4+(20)/(5+4x-x^2)]dx`

Then apply basic integration property: `int (u+-v) dx= int(u) dx +- int (v) dx`

where we can integrate each term separately.

`int [-x-4+(20)/(5+4x-x^2)]dx=int (-x) dx -int 4dx +int (20)/(5+4x-x^2)]dx`

For the integration of `int (-x) dx` , we may apply Power Rule integration: `int u^n du= u^(n+1)/(n+1)+C`

`int (-x) dx = - int x dx`

`=-x^(1+1)/(1+1)`

`= -x^2/2`

For the integration of` -int 4 dx` , we may apply basic integration property:`int c*f(x)dx= c int f(x) dx`

`-int 4dx = -4 int dx`

`= -4x`

For the integration of `int 20/(5+4x-x^2)dx` , we apply partial fractions:

`(20)/(5+4x-x^2) = 20/(6(x+1)) -20/(6(x-5))`

Then, `int (20)/(5+4x-x^2)dx=int [20/(6(x+1)) -20/(6(x-5))]dx`

Apply** basic integration property**: `int (u+-v) dx= int(u) dx +- int (v) dx`

`int [20/(6(x+1)) -20/(6(x-5))]dx =int 20/(6(x+1))dx-int 20/(6(x-5))dx`

Apply apply the basic integration property: `int c*f(x)dx= c int f(x) dx` and **basic integration formula for** **logarithm**: `int (du)/u = ln|u|+C` .

`int 20/(6(x+1))dx-int 20/(6(x-5))dx =(20/6)int 1/(x+1)dx-(20/6)int 1/(x-5)dx`

`=(20/6)ln|x+1|- (20/6)ln|x-5|`

For the right side, we get:

`int [-x-4+(20)/(5+4x-x^2)]dx= -x^2/2-4x+(20/6)ln|x+1|- (20/6)ln|x-5|+C`

Note: Just include the constant of integration "C" on one side as the arbitrary constant of a differential equation.

Combining the results from both sides, we get the **general solution of the differential equation**:

`y =-x^2/2-4x+(20/6)ln|x+1|- (20/6)ln|x-5|+C`

`y =-x^2/2-4x+(10/3)ln|x+1|- (10/3)ln|x-5|+C`