∫1/√(1+x^2) dx

t = x + √(1+x^2)
t - x = √(1+x^2)

∫1/√(1+x^2) dx = ∫1/(t-x) dx　・・・[1]

また、dt/dx = 1 + x/√(1+x^2) = (x + √(1+x^2))/√(1+x^2) = t/(t-x)
∴dx/dt = (t-x)/t

∴[1] = ∫(1/(t-x))*(t-x)/t dt = ∫(1/t) dt = log| t | + C = log| x+√(1+x^2) | + C