>>741
∫_[2〜3] √(1＋y'^2) dx＝∫_[2〜3] √(1＋((1＋x/√(x^2−1))/(x＋√(x^2−1)))^2) dx
＝∫_[2〜3] x/√(x^2−1) dx
x＝cosh(t)
∫_[2〜3] x/√(x^2−1) dx＝∫_[x=2〜3] cosh(t) dt＝[sinh(t)]_[x=2〜3]＝[√(x^2−1)]_[2〜3]＝√8−√3