>>93
y = √(ax^2+bx+c)
t = (√a)x+√(ax^2+bx+c)とおく
ax^2+bx+c = t^2 + ax^2 - 2t(√a)x
t^2 = (b+2t(√a))x + c
x = (t^2-c)/(b+2(√a)t)
dx = (2t(b+(2√a)t)-(t^2-c)(2(√a)))/(b+2(√a)t)^2 dt
= (2bt+4(√a)t^2-2(√a)t^2+2c√a)/(b+2(√a)t)^2 dt
= 2((√a)t^2+bt+(√a)c)/(b+2(√a)t)^2 dt

y = t-(√a)x
= t-(√a)(t^2-c)/(b+2(√a)t)
= (t(b+2(√a)t)-(√a)(t^2-c))/(b+2(√a)t)
= (bt+2(√a)t^2-(√a)t^2+(√a)c)/(b+2(√a)t)
= ((√a)t^2+bt+(√a)c)/(b+2(√a)t)

∫ydx = ∫(((√a)t^2+bt+(√a)c)/(b+2(√a)t) * (b+2(√a)t)^2/2((√a)t^2+bt+(√a)c)) dt
= ∫(b+2(√a)t)/2 dt
= ∫(b/2+(√a)t)dt
= 1/2(bt+(√a)t^2)

ああ面倒くせえ。あとは自分で代入してくれ