∫ln(x+√(1+x^2))dx
=xln(x+√(1+x^2) -∫xd(ln(x+√(1+x^2))
[ln(x+√1+x^2)]'=[1+x/√(1+x^2)]/(x+√(1+x^2))=1/√(1+x^2)
=xln(x+√(1+x^2)-∫xdx/√(1+x^2)
=xln(x+√(1+x^2)-(1/2)∫d(1+x^2)/√(1+x^2)
=xln(x+√(1+x^2)-√(1+x^2)+C不好意思，没有ln