请问这2题定积分 如何解1.∫x arctanx dx上限是1下限是02.∫sin(lnx)dx上限是e^(x/2)下限

1个回答

  • 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