∫x·ln(1+X的2次方)dx
=0.5∫ln(1+X²)d(1+x²)
设t=1+X²
=0.5∫lntdt
=0.5tlnt-0.5∫t*(1/t)dt
=0.5tlnt-0.5t+C
=0.5(1+x²)ln(1+x²)-0.5(1+x²)+C