∫lnxdx/(1+x^2)^(3/2)=(x=tanu代换)=∫lnxd[x/(1+x^2)^(1/2)]=xln[x/(1+x^2)^(1/2)]-∫dx/(1+x^2)^(1/2)x=tanu代换=xln[x/(1+x^2)(1/2)-1/2ln[(1+sin(arctanx)/(1-sinarctanx)] sin(arctanx)=x/(1+x^2)^(1/2)
