09/04/02 03:07:45
めんどくさいけど、そこまで悩む問題でもない
∫[√3, 2√2] √(1+(1/x^2)) dx
= ∫[√3, 2√2] (√(x^2+1))/x dx
x = tan(θ) と置換
= ∫[π/3, arctan(2√2)] dθ/(sin(θ)cos^2(θ))
cos(θ) = t と置換
= ∫[1/3, 1/2] dt/(t^2(1-t^2))
= (1/2) ∫[1/3, 1/2] {1/(1+t) + 1/(1-t) + 2/t^2} dt
= (1/2) [log((1+t)/(1-t)) - 2/t]_[1/3, 1/2]
= (1/2)log(3/2) + 1