Solution;
dvdu=(u−v)2−2(u−v)−2 
 Take ;
x=(u−v) 
Such that;
dvdx=dvdu−1 
Substitute into the equation;
dvdx+1=x2−2x−2 
dvdx=x2−2x−3 
Seperate by variables;
x2−2x−31dx=dv 
Integrating;
4ln(∣x−3∣)−ln(∣x+1∣)=v+c 
But ;
x=u-v;
Replace back;
4ln(∣u−v−3∣)−ln(∣u−v+1∣)=v+C 
                             
Comments