∫1/x(x^2+1)^(1/2)dx=?请朋友们说下过程吧,
问题描述：
∫1/x(x^2+1)^(1/2)dx=?请朋友们说下过程吧,
答
答：
设t=√(x^2+1)，x=√(t^2-1)
∫ {1/[x√(x^2+1)]} dx
=∫ {1/[t*√(t^2-1)] d[√(t^2-1)]
=∫ {1/[t*√(t^2-1)]}*(1/2)*[2t/√(t^2-1)] dt
=∫ [1/(t^2-1)] dt
=(1/2) ∫ [1/(t-1)-1/(t+1)] dt
=(1/2)*[ln|t-1|-ln|t+1|+C
=(1/2)*ln|(t-1)/(t+1)|+C
=(1/2)*ln|[√(x^2+1)-1]/[√(x^2+1)+1]|+C
答
∫ dx/[x√(x^2+1)]
let
x= tany
dx = (secy)^2dy
∫ (secy/tany) dy
=∫ cscy dy
=ln|cscy-coty| + C
=ln|√(x^2+1)/x - 1/x | + C