>>96
∫[x=1〜a] √(x^2+1)/x dx、√(x^2+1)=tとおくと、∫[t=√2〜√(a^2+1)] 1 + (1/2){1/(t-1) - 1/(t+1)} dt
={t + (1/2){log|t-1)/(t+1)|}_[t=√2〜√(a^2+1)]=√(a^2+1)-√2+log{(√(a^2+1)-1)/a}-log(√2-1)