1.本题可对x分部积分:
∫x arctanx dx=x^2arctanx| +∫x^2/(1+x^2) dx
=π/4+1-∫1/(1+x^2) dx
=π/4+1-π/4=1
2.本题换元,再分部积分即可;
令t=lnx代入,上限不知道是不是e的根号2次方.则上下限变为根号2和0.
∫sin(lnx)dx=∫sin(t)e^(t)dx=I
对e^(t)分部积分:
=∫sin(t)de^(t)
=sin(t)e^(t)| +∫cos(t)e^(t)dx
(在分部积分)=sin(根号2)+cos(t)e^(t)| -∫sin(t)e^(t)dx
=sin(根号2)+cos(根号2)-I
2×I=sin(根号2)+cos(根号2)
I=[sin(根号2)+cos(根号2)]/2=∫sin(lnx)dx