求解考研数学积分

= x * ln(1+√( 1+1/x)+1/2∫1/x * 1/[1+1/x+√( 1+1/x)]dx

设(1+1/x)= T2 t > 0 x=1/(t^2-1)dx=-2t*(t^2-1)^-2

代入1/2∫1/x * 1/[1+1/x+√( 1+1/x)]dx。

=-∫dt/(t+1)(t^2-1)=∫1/(t+1)+2/(t+1)^2-1/(t-1)dt

= ln |(t+1)/(t-1)-2/(t+1)(1+1/x)=t^2

因此

= x * ln(1+√( 1+1/x)+ln |(t+1)/(t-1)-2/(t+1)√(1+1/x)= t