做兩題吧,用不定積分方法
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
可得定積分為4pi/3-3^(1/2)
3.令t=1/x 則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
可得定積分為pi/2-arcsin(3/4)