求定积分∫arctanxdx 上限是1 下线是6 请高手帮忙谢谢 急

∫arctanxdx=xarctanx-∫xd(arctanx)=xarctanx-∫x/(1+x^2)dx=xarctanx-1/2∫dx^2/(1+x^2)=xarctanx-1/2ln(1+x^2)+C
分步积分
代入上下限，答案为π/4-1/2ln2-(6arctan6-1/2ln37)=π/4-6arctan6-1/2ln2/37

∫arctanxdx
=xarctanx-∫xdx/(1+x^2)
=xarctanx-(1/2)∫d(x^2+1)/(1+x^2)
=[xarctanx-ln(1+x^2)/2]|带入
=派/4-(ln2)/2-6arctan6+(ln37)/2
=派/4-6arctan6+(ln37/2)/2

1/(1+x^2)自己代入入