∫[x=1〜2] √{1+(1/x^2)} dx = ∫[x=1〜2] √(x^2+1)/x dx、√(x^2+1)=t とおくと、
dx={√(x^2+1)/x}dt より、
∫[t=√2〜√5] 1 + 1/(t^2-1) dt = ∫[t=√2〜√5] 1 + (1/2){(1/t-1) - (1/t+1)} dt