答:
即f'(x)=xln(1+x²)
所以f(x)=∫xln(1+x²)dx
=x²ln(1+x²)/2-∫x³/(1+x²)dx
=x²ln(1+x²)/2-∫(x³+x-x)/(1+x²)dx
=x²ln(1+x²)/2-∫x-x/(1+x²) dx
=x²ln(1+x²)/2-x²/2+ln(1+x²)/2 + C
=(x²+1)ln(1+x²)/2-x²/2 + C
因为y=f(x)过(0,1/2)
所以1/2=(0+1)ln(1+0)/2-0/2+C,即C=1/2
所以f(x)=(1+x²)ln(1+x²)/2+(1-x²)/2