原式=1/2∫(0→1)arctanxd(x^2)
=1/2x^2arctanx|(0→1)-1/2∫(0→1)x^2*1/(1+x^2)dx
=1/2x^2arctanx|(0→1)-1/2∫(0→1)(x^2+1-1)/(x^2+1)dx
=1/2x^2arctanx|(0→1)-1/2∫(0→1)dx+1/2∫(0→1)dx/(x^2+1)
=1/2x^2arctanx|(0→1)-x/2|(0→1)+1/2arctanx|(0→1)
=π/8-1/2+π/8
=π/4-1/2
求定积分上限1下限0xarctanxdx