∫ dx/√(x^2+a^2) 令 x = a tant,dx = a(sect)^2 dt,√(x^2+a^2) = a sect
= ∫ sect dt
= ln(sect + tant| + C1
= ln(x+√(x^2+a^2)) + C1- lna
= ln(x+√(x^2+a^2)) + C∫ sect dt = ln(sect + tant| + C1如何得出的。∫ sect dt = ∫ 1/cost dt = ∫ cost /[1-(sint)^2] dt u = sint= ∫ 1/(1-u^2) du =(1/2)∫ [ 1/(1-u) + 1/(1+u) ] du= (1/2) ln (1+u)/(1-u)+ C = (1/2) ln [(1+u)^2/(1-u^2)] + C= ln |(1+sint)/cost| + C = ln |sect + tant | + C以后这就是需要背下来的基本公式了。