∫xln(1+x^2)dx
=(1/2)∫ln(1+x^2)d(x^2)设x^2=u
=(1/2)∫ln(1+u)du
=(1/2)[uln(1+u)-∫u/(1+u)du]
=(1/2)[uln(1+u)-∫1-1/(1+u)du]
=(1/2)[uln(1+u)-u-ln(1+u)]+C转换回去
=(1/2)[x^2ln(1+x^2)-x^2+ln(1+x^2)]+C