08/05/18 13:00:36
訂正：
→ ∫(1-2u)/(u^2-u+1)du=2∫dx/(x+1)
→ -log|u^2-u+1|=2log|x+1|+C → 1/(u^2-u+1)=C(x+1)^2
→ (x+1)^2-(x+1)(y-1)+(y-1)^2=x^2+y^2-xy+3x-3y+3=C → x^2+y^2-xy+3x-3y=C
08/05/18 13:00:36
訂正：
→ ∫(1-2u)/(u^2-u+1)du=2∫dx/(x+1)
→ -log|u^2-u+1|=2log|x+1|+C → 1/(u^2-u+1)=C(x+1)^2
→ (x+1)^2-(x+1)(y-1)+(y-1)^2=x^2+y^2-xy+3x-3y+3=C → x^2+y^2-xy+3x-3y=C