求定积分 ∫arcsin根号(x/(1+x)dx 等

2个回答

  • 做两题吧,用不定积分方法

    1.∫arcsin[x/(x+1)]^(1/2)dx

    =xarcsin[x/(x+1)]^(1/2)

    -∫x/[1-x/(x+1)]*(1/2)*[(x+1)/x]^(1/2)*(x+1)^(-2)dx

    =xarcsin[x/(x+1)]^(1/2)-∫x^(1/2)/2(x+1)dx

    =xarcsin[x/(x+1)]^(1/2)-∫1/2x^(1/2)-1/2x^(1/2)*(x+1) dx

    =xarcsin[x/(x+1)]^(1/2)-x^(1/2)+arctan[x^(1/2)]+C

    可得定积分为4pi/3-3^(1/2)

    3.令t=1/x 则dx=-dt/t^2

    ∫dx/x(3x^2-2x-1)^(1/2)

    =∫-(dt/t^2)*t|t|/(3-2t-t^2)^(1/2)

    =-sgn(t)∫dt/[4-(t+1)^2]^(1/2)

    =-sgn(t)arcsin[(t+1)/2]+C

    =-sgn(x)arcsin[(x+1)/2x]+C

    可得定积分为pi/2-arcsin(3/4)