∫ln(1+x^1/2)DX 用分部积分法求解不大明白,请老大们把步骤及答案写出来,

分部积分法：∫f(x)dg(x)=f(x)g(x)-∫g(x)df(x)先做个变量置换,令x^1/2=t则∫ln(1+x^1/2)dx=∫ln(1+t)d(t^2)=ln(1+t)(t^2)-∫(t^2)dln(1+t)=ln(1+t)(t^2)-∫(t^2)/(1+t)dt=ln(1+t)(t^2)-∫(t^2-1)/(1+t)dt+∫1/(1+t)...