Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 89

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